Download Chapter 0: Prologue

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Chapter 0: Prologue
Surprising Results
Frequently a course in high school geometry starts with a long list of definitions and proofs
of theorems which are intuitively obvious. It rarely includes theorems and proofs which
make a lasting impression on the student.
Pause for a moment and try to list some geometrical theorems you can recall. Are there
any on your list that contain an unexpected or surprising result? Which are especially
beautiful? Are any of them especially important and if so why?
This short section introduces some problems and theorems from Euclidean geometry
whose solutions or statement you will very likely find surprising. In several cases, you
will be asked to conjecture a solution or a theorem through a sequence of suggested
experiments. To do this you will need a compass, ruler, and some blank sheets of paper.
A geometry utility software such as GSP (Geometer’s Sketchpad) is especially helpful in
investigating geometrical properties and making conjectures. In Appendix II, we introduce
the basics of GSP and throughout the text in the margins suggest optional activities using
the software whenever appropriate. However the text is independent of GSP and can be
read and studied without the software. Later in this text you will learn how to prove some
of the conjectures and statements. Several of the proofs are particularly beautiful due to
their unexpected simplicity and applicability to new problems.
1.
The Treasure Island Problem
Among his great-grandfathers’s papers Marco found a parchment describing the location
of a hidden pirate treasure buried on a desert island. The island contained a coconut tree,
a banana tree, and a gallows (Γ) where traitors were hanged. A reproduction of the map
appears in Figure 0-1 and was accompanied by the following directions:
“ Walk from the gallows to the coconut tree, counting the number of steps. At
the coconut tree, turn 90◦ to the right. Walk the same distance and put a spike
in the ground. Return the gallows and walk to the banana tree counting your
steps. At the banana tree turn 90◦ to the left and walk the same number of
steps and put another spike in the ground. The treasure is halfway between the
spikes.”
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Figure 0-1
GSP 0-1
Marco found the island and the two trees but no trace of the gallows or the spikes
which had probably rotted. In desperation Marco began to dig at random but soon gave
up because the island was too large. Try to devise a plan for finding the treasure.
If you have spent enough time pondering a solution but could not find one, try the
following: on a piece of paper fix the positions of the two trees. Choose an arbitrary
position for the gallows and mark it Γ1 . Follow the directions to find the corresponding
spikes and the midpoint between them T1 . Next choose another position for the gallows
Γ2 and follow the directions to find the corresponding location of the treasure T2 . Repeat
the procedure for at least two more positions of the gallows. What do you notice about
T1 , T2 , T3 and T4 ? Now try to conjecture how to locate the treasure.
2.
Circle Rolling Around a Circle
In Figure 0-2, a quarter is shown atop another quarter each with head up. If the top
quarter is rolled around the bottom one until it rests directly below the bottom quarter,
will the head be straight up or upside down? Answer the question based on your intuition
and then check your conjecture using two actual coins.
Figure 0-2
A related problem involves rolling a circle around a different size stationary circle. In
particular suppose that the radius of circle A is 1/3 the radius of circle B. Starting at
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point P shown in Figure 0-3, circle A rolls all the way around circle B until it returns
back to point P . How many revolutions about its own center does circle A make? Guess
the answer and then check it by performing an appropriate experiment. You may also try
to figure out the number of revolutions if circle A rolls inside circle B and finally if B rolls
around circle A. These questions will be answered in later chapters. Again several proofs
will be given based on the material studied.
Figure 0-3
3.
The Nine-Point Circle
The 19th century experienced a renewed interest in classical Euclidean geometry. Probably the most spectacular discovery was the Nine-Point Circle which was investigated
simultaneously by the French mathematicians Charles Jules Brianchion (1785-1864) and
Jean-Victor Poncelet (1788-1867) and published jointly in 1821. However, the theorem is
commonly attributed to the German mathematician and high school teacher Karl Wilhem
Feurbach (1800-1834) who independently discovered the theorem and published it and
some related results in 1822.
With any triangle ABC, nine particular points can be associated as shown in Figure
0-4: the midpoints of the three sides, M1 , M2 , M3 ; the midpoints of the segments joining
the vertices A, B, C and H, the point of intersection of the altitudes, N1 , N2 , N3 (it will
be proved in Chapter 1 that the three altitudes of a triangle intersect in a single point);
and the feet of the three altitudes F1 , F2 , F3 (the points where the altitudes intersect
corresponding sides).
The theorem states that all nine points lie on one circle called the Nine-Point Circle.
Further properties of the nine-point circle will be discussed in the problem set. The NinePoint Circle Theorem and related results will be proved in Chapter ??.
GSP 0-2
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Figure 0-4
4.
GSP 0-3
Morley’s Theorem
In 1899 Frank Morley, professor of mathematics at Haverford College and later at John
Hopkins University discovered an unusual property of the trisectors of the three angles of
any triangle.
If the angle trisectors are drawn for each angle of any triangle, then the adjacent
trisectors of the angles meet at vertices of an equilateral triangle. In Figure 0-5 the
adjacent trisectors of angles A and B meet at D, the adjacent trisectors of angles A and
C meet at E, and the adjacent trisectors of angles C and B meet at F . Morley’s Theorem
states that triangle F DE is equilateral (i.e. its three sides are all the same length). Since
1899, many different proofs of the theorem have been published, several as late as the
1970’s. A proof of Morley’s theorem will be presented in Chapter 4.
Figure 0-5
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5.
The Hiker’s Path
A hiker H in Figure 0-6 need to get to the river r and then to her tent T . Find the point
X on the bank of the river so that the hiker’s total trip HX + XT is as short as possible.
GSP 0-4
Figure 0-6
6.
The Shortest Highway
A highway connecting two cities A and B as in Figure 0-7 needs to be built so that part
of the highway is on a bridge perpendicular to the parallel banks b1 and b2 , of a river.
Where should the bridge be built so that the path AXY B is as short as possible?
Figure 0-7
7.
Steiner’s Minimum Distance Problem
One of the greatest geometers of all time, and certainly of the 19th century, was Jacob
Steiner (1796-1863). Born in Switzerland but educated in Germany, Steiner discovered and
GSP 0-5
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proved new theorems and introduced new geometrical concepts (e.g. inversive geometry
which will be discussed later in the text). In particular, Steiner was interested in the
solutions of maximum and minimum problems using only purely geometric methods, that
is, without using calculus or algebra. Among others he proved the following theorem
illustrated in Figure 0-8.
Figure 0-8
GSP 0-6
If A, B and C are three points forming a triangle such that each of the angles of
∆ABC is less than 120◦ , then the point for which the sum of the distances from P to the
vertices of the triangle (that is P A + P B + P C) is minimum has the property that each
of the angles at P is 120◦ . Steiner also dealt with the case when one of the angles is 120◦
or greater that 120◦ . In addition, he generalized the problem for n points.
Many maximum and minimum problems can be more efficiently solved using purely
geometrical methods rather than calculus. The following two problems are examples of
such problems that will be solved using transformational geometry in Chapter 5.
8.
The Pythagorean Theorem
One of the best known and most useful theorems in geometry and perhaps all of mathematics is the Pythagorean Theorem discovered in the sixth century B.C.E. There are
numerous known proofs of the theorem (The Pythagorean Proposition by E. Loomis contains hundreds of them). Do you recall any proof of the Pythagorean Theorem? Can you
prove it on your own?
The following is one way to state the theorem: If squares are constructed on the sides
of any right triangle (a triangle with a 90◦ angle), then the area of the largest square equal
the sum of the area of the other squares. In Figure 0-9, if the areas of the squares are A,
B, and C, then A + B = C or, equivalently, a2 + b2 = c2 , where a and b are the lengths of
the legs of the triangle and c is the length of the hypothenuse (the side opposite the right
angle).
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Figures 0-9 and 0-10 can be used to justify the Pythagorean Theorem. Can you see
how?
Figure 0-9
Figure 0-10
In later chapters, we will give several proofs of the Pythagorean Theorem and investigate what other figures can be constructed on the sides of a right triangle so that the area
of the figure constructed on the hypothenuse equals the sum of the areas of the figures constructed on the other two sides. We will also generalize the theorem in three-dimensional
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space and for triangles which are not necessarily right triangles.
Problem Set 0
In each of the following problems you may use any tools to perform the experiments (a
geometry drawing utility such as The Geometer’s Sketchpad is especially convenient but
not necessary).
1. (a) Conjecture the solution to the Treasure Island Problem by choosing at least
four different positions for the gallows as described in the text.
(b) Does your conjecture hold for some Γ positioned below the line connecting the
coconut and banana tree? (Make an appropriate construction)
(c) Choose Γ at one of the trees and find the corresponding treasure.
(d) Choose Γ on the line connecting the trees half way between them and find the
corresponding treasure.
(e) Based on your experiments in (a) through (d) what seems to be the simplest
way to find the treasure?
2. (a) Draw a circle on transparent (or see through) paper. How would you find the
center of the circle by folding the circle onto itself?
(b) Draw an arc of a circle. How could you find the center now?
3. Let ABCD be any convex quadrilateral. On each side of the quadrilateral construct
a square as shown. Find the centers C1 , C2 , C3 , and C4 of the squares, where C1 and
Figure 0-11
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C3 are centers of opposite squares. How are the segments C1 C3 and C2 C4 related?
Repeat the experiment starting with a different quadrilateral.
4. Choose an arbitrary triangle ABC and construct the corresponding Nine-Point Circle. (You may need the result from problem 2.)
5. Check Morley’s Theorem experimentally for some triangle ABC.
6. Use Figure 0-10 to prove the Pythagorean Theorem.
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Chapter 1: Congruence,
Constructions, and the Parallel
Postulate
At the age of eleven, I began Euclid. with my brother as my tutor. This was one of the
great events of my life, as dazzling as first love. I have not imagined there was anything
so delicious in the world . . . From that moment until . . . I was thirty-eight, mathematics
was my chief interest and my chief source of happiness.
- Bertrand Russell(1872-1970)
The Autobiography of Bertrand Russell
London: G. Allen and Unwin, 1968
HISTORICAL NOTE: Bertrand Russell
Lord Bertrand Arthur William Russell was a British philosopher, logician and
mathematician who made important contributions to foundations of mathematics.
In 1910, he became a lecturer at Cambridge University but was dismissed and later
jailed for making pacifist speeches during World War I. He abandoned pacifism during
World War II in the face of the Nazi atrocities in Europe and Nazi threat to Great
Britain and . After the war he reverted back to pacifism, and became a leader in
the anti-nuclear and anti-Vietnam War movements. Later Russell taught at several
United States universities including Harvard and the University of Chicago. He won
the Noble Prize for Literature in 1950. Russell died at the age of 98.
Introduction
In chapter we lay the foundation for Euclidean and Non-Euclidean Geometry. About 250
B.C.E. Euclid systematically collected and organized the geometrical knowledge of his
time in a treatise composed of thirteen books, The Elements. (The treatise also included
number theory and topics in algebra.) Euclid started out with a list of statements called
axioms or postulates that he assumed to be true and showed that geometric statements
follow logically from his assumptions. However, he did not realize that some geometric
terms could not be defined. Some terms must be left undefined in order to avoid “circular”
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definitions. (For example, in one dictionary a line is defined as “the path traced by a moving
point” and then a path is defined as “a line obtained by a moving point.”) Euclid defined
a point as that which has no part and a line as breadthless length. Neither part, nor
breadth, nor length has been defined.
In his proofs Euclid used unstated assumptions and relied on diagrams to make what
seemed to him to be obvious conclusions. In spite of the above shortcomings, Euclid’s
achievements have been monumental. In presenting a vast amount of mathematics by
starting with a few basic assumptions and logically deducing other mathematical statements, Euclid laid the foundation for a deductive approach to mathematics. Only towards
the end of the nineteenth century was a rigorous foundation of Euclidean Geometry established. In 1899 David Hilbert with his book Foundations of Geometry established a set
of axioms along with undefined terms for Euclidean Geometry and succeeded in proving
Euclid’s theorems relying only on logic. Hilbert’s success had roots in a revolution that
took place in geometry earlier. In 1829 the Russian mathematician Nikolai Lobachevsky
and independently two years later the Hungarian mathematician Janos Bolyai established
a new geometry referred to as Non-Euclidean Geometry. At the same time the great German mathematician Karl Friedrich Gauss was very likely aware of the new results, but
did not publish them as he was worried about the controversy that they might arouse.
Non-Euclidean geometry is based on an axiom that denies Euclid’s Fifth Postulate
(Euclid used the term postulate for geometric assumptions and axiom for general mathematical non-geometrical assumptions. For example, one of Euclid axioms is: ”The part is
smaller that the whole” and one of his postulates: ”A straight line segment can be drawn
joining any two points.”) Euclid’s Fifth Postulate or the Parallel Postulate is equivalent
to the following:
Euclidean Parallel Postulate Through a given point P not on a line ! there is
exactly one line parallel to !.
Figure 1-1
For generations mathematicians believed it was possible to prove the parallel postulate
using Euclid’s other postulates or axioms and tried to do so. Only Lobachevsky, Bolyai,
and Gauss were bold enough to substitute Euclid’s Parallel Postulate by another one which
denied Euclid’s. It is referred to as the Hyperbolic Parallel Postulate.
Hyperbolic Parallel Postulate: Given a line ! and a point P not on ! there exists
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at least two lines through P which are parallel to !.
Euclid’s approach to geometry was directed by physical reality. Thus points and lines
were idealized mathematical terms for what we perceive by a point and straight line.
Postulates were “self-evident” truths. To the originators of Hyperbolic Geometry, points
and lines were undefined terms satisfying certain axioms as such they do not necessarily
represent any idealized physical objects. (We will see in later chapters that it is possible
to find familiar objects to represent points, lines, and planes which satisfy the Hyperbolic
Parallel Postulate.) We will discuss Hyperbolic Geometry in later chapters.
In Sections 1.2 and 1.3 we develop Neutral Geometry (also called Absolute Geometry)
in which the Parallel Postulate is not used. Consequently, theorems in Neutral Geometry
are true in Euclidean as well as Hyperbolic Geometry. We adopt a set of axioms which is
a modification of the axioms introduced by the American mathematician George David
Birkoff (1884-1944). The axioms are consistent and independent. A set of axioms is
consistent if no contradictions can be derived from the set and independent if no axiom
of the set is implied by the other axioms in the set.
In order to get quickly to interesting results we placed the axioms and related material
in Appendix I.
HISTORICAL NOTE: Euclid
Euclid (Third Century B.C.E.) The place and exact year of Euclid’s birth are not
known. Historians believe he was the first mathematics professor at the University
of Alexandria. Euclid’s The Elements was the first treatise on mathematics as a
deductive system. Except for the Bible it has been more widely studied than any
other book. Over a thousand editions of The Elements have appeared since its first
printing in 1482. However, no Greek copy from Euclid’s time has been found. The
Elements reached the West from an Arabic translation. The teaching of geometry has
been dominated by The Elements for over 20 centuries. The Elements has also had a
profound influence on scientific thinking.
1.1
Angles and Their Measurement
Pairs of Angles
Certain pairs of angles occur often enough that it is convenient to give them special names.
Two angles are adjacent if they lie in the same plane, share a common side and the
interior of the angles have no point in common. ∠CAD and ∠DAB in Figure 1-2 are
adjacent angles. The common side of two adjacent angles of equal measure (in this text
we mostly measure angles in degrees) is called an angle bisector.
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Figure 1-2
Two angles are a linear pair if they are adjacent and the noncommon sides are
opposite rays. (See Figure 1-3 (a).)
Two angles are vertical if their sides form two pairs of opposite rays. (See Figure 1-3
(b).)
(a)
(b)
Figure 1-3
It seems obvious and can be readily deduced that the sum of the measures of two
angles in a linear pair is 180◦ . We often encounter pairs of angles whose measures add
up to 180◦ or 90◦ . Such angles have a special name. Two angles are supplementary if
the sum of their measure is 180◦ . Each angle is called the supplement of the other. Two
angles are complementary if the sum of their measures is 90◦ . Each angle is called the
complement of the other.
Notice that two angles in a linear pair are supplementary but not every two supplementary angles form a linear pair. In fact it follows that if two angles are supplementary
and adjacent then they form a linear pair. An angle measuring less than 90◦ is acute and
one measuring more than 90◦ is obtuse.
It is common to call a 90◦ angle a right angle. An angle measuring 180◦ is a straight
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angle. Congruent angles are angles which have the same measure (we denote the
measure of an angle by m). Thus if m(∠A) = m(∠B), ∠A and ∠B are congruent. We
use the symbol ∼
= for congruent. Hence ∠A ∼
= ∠B. Notice that the equality symbol “ = ”
is reserved for real numbers or for sets which have the same elements (the degree measure
is a real number). Because two angles which have the same measure are not necessarily
the same set of points, we do not use the equality symbol between congruent angles.
Notation for Angles and Their Measures
It is often cumbersome to use the notation m(∠A) or m(∠BAC) for the measure of an
angle. We commonly use small Latin or Greek alphabet letters for measures of angles.
Figure 1-4
In Figure 1-4 we designated the measures of the three angles of "ABC by α, β and
γ (notice the correspondence to A, B, and C). Thus m(∠A) = α, m(∠B) = β, and
m(∠C) = γ. (Because α is already the measure of ∠A, we do not write m(α) or m(∠α).)
The following theorems follow immediately from the above definitions. We prove the
second theorem.
Theorem 1.1.
(a) Supplements of congruent angles are congruent.
(b) Complements of congruent angles are congruent.
(c) The sum of the measures of two angles in a linear pair is 180◦ .
Theorem 1.2. Vertical angles are congruent.
Proof. Designating the measures of the angles in Figure 1-5 by α, β, γ, and δ, we need
to show that α = β and γ = δ. Notice that α + γ = 180 and β + δ = 180, which implies
that α = β. (Alternately we could say that α is a supplement of β and γ is a supplement
of δ and hence by Theorem 1.1, α = β.)
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Figure 1-5
When two lines intersect they form four angles as shown in Figure 1-5. If one of the
angles formed is a right angle the lines are perpendicular. Previous theorems imply
that in this case all four angles are right angles. The fact that lines are perpendicular
or equivalently that an angle formed is a right angle is designated in drawings by the
symbol " as shown in Figure 1-6. We also write a⊥b to designate that lines a and b are
perpendicular.
Figure 1-6
When a line intersects a segment at its midpoint and is perpendicular to the segment
it is called the perpendicular bisector of the segment. The line ! in Figure 1-7 is the
perpendicular bisector of AB. Notice that M is the midpoint and the congruent segments
AB and M B are marked by the same symbol.
Figure 1-7
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−−→
−−→
Example 1.1. In Figure 1-8, ∠AOB and ∠BOC are a linear pair. The rays OD and OE
←→ ←→
are their angle bisectors respectively. Prove that OE⊥OD.
Figure 1-8
−−→
−−→
Solution. Because OD and OE are angle bisectors two pairs of congruent angles are
formed. We designate the measures of the congruent angles in each pair by α and β
respectively as shown in Figure 1-8. Because ∠AOB and ∠BOC are a linear pair we have:
2α + 2β = 180◦ . Hence α + β = 90◦ .
←→ ←→
Because m(∠EOD) = α + β, it follows that m(∠EOD) = 90◦ and hence OE⊥OD.
Problem Set 1.1
1. OA⊥OC and OB⊥OD. Which of the non right angles formed are congruent? Justify
your answer.
2. Examine the following argument showing that the sum of the measures of the angles
in any triangle is 180◦ . Is the argument valid? Justify your answer.
In "ABC we choose an arbitrary point D in the interior of the triangle and connect
it to the three vertices of the triangle. We then mark the measures of the nine
angles as shown in the figure and let the sum of the measures of the three angles in
a triangle be k. We show that k = 180.
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Adding up all the measures of the angles in the three triangles "ABC, "ACD and
"CBD we get 3k. Hence: (x+u+r)+(v +z +t)+(w +y +s) = 3k. By commutative
and associative properties of addition we get (x + u + v + z + w + y) + (r + t + s) = 3k.
Notice that the quantity in the first parenthesis represents the sum of the measures
of the angles in "ABC and hence equals k. This and the fact that r + t + s = 360
imply that: k + 360 = 3k or k = 180.
3. Jaimee announced that she has her own somewhat different proof of Example 1.1.
Examine Jaimee’s proof and compare it with the one given in the text. Answer the
following:
a. Discuss strengths and any weaknesses exposition in Jaimee’s exposition? In
case of shortcomings suggest improvements.
b. Which proof do you like better? Why?
Jaimee:
I know that if a ray divides a straight angle into two congruent angles, each must be
90◦ . Thus, to show that EO ⊥ OD, I need only to prove that ∠F OE = ∠EOD. I
extend the rays in Figure 1.1 and use the fact that the bisector of an angle divides
it into two congruent angles, and the fact that vertical angles are congruent. I mark
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the angles by x or by y as shown in my figure above. ∠F OD is a straight angle
because I extended the ray OD. I see that ray OE divides ∠F OD into two angles,
each equal to x + y. Hence each angle is 90◦ .
1.2
Congruence of Triangles
Informally when two figures have the same size and shape they are congruent. In Section 1.2 we have defined congruent segments as segments which have the same length, and
angles to be congruent if they have the same measure. When two figures are congruent it
is always possible to fit one figure onto the other so that matching sides and angles are
congruent. This is the basis for defining congruent triangles.
(a)
(b)
Figure 1-9
Definition of Congruent Triangles
Triangles are congruent if there is a one-to-one correspondence between their
vertices so that corresponding sides are congruent and corresponding angles
are congruent.
In Figure 1-9, "ABC and "DEF are congruent. Corresponding sides and angles
are marked as shown. Notice that if we fit one triangle onto the other, vertex A will
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correspond to vertex D, B to E, and C to F (Try to superimpose one onto the other by
tracing "DEF on a sheet of paper. You will meed to flip the sheet and trace the triangle
on the other side of the paper before superimposing it on "ABC). The fact that the
triangles are congruent is written "ABC ∼
= "DEF . Whenever the symbol “∼
=” is used,
corresponding vertices must be written in the same position. Thus without reference to
a figure the congruence "ABC ∼
= "DEF implies ∠A ∼
= ∠D, ∠B ∼
= ∠E, ∠C ∼
= ∠F ,
∼
∼
∼
AB = DE, BC = EF , and AC = DF .
If three sides and three angles of one triangle are congruent to corresponding sides
and angles of another triangle, by definition the triangles are congruent. It turns out that
congruence of fewer corresponding parts is sufficient to determine that two triangles are
congruent.
Figure 1-10 shows "ABC and "DEF in which AB ∼
= DE, ∠A ∼
= ∠D, AC ∼
= DF .
(a)
(b)
Figure 1-10
It seems that these conditions assure that the triangles are congruent. This can be
seen as follows. Because ∠A ∼
= ∠D we could superimpose ∠D on top of ∠A so that ray
−−→
−−→
−−→
−→
DE falls on ray AB and DF falls on AC. Because AB ∼
= DE and AC ∼
= DF the vertex
E will fall on B and F on C. Thus the vertices D, E, F will fall on the corresponding
vertices A, B, C and therefore the triangles are congruent.
However, notice that the above argument is an intuitive justification and not a proof.
We don’t have a rigorous definition of what it means to superimpose one figure on top of
another. Later in Chapter 6 during the study of transformations we will be able to give a
precise definition of what it means to move a figure. Meanwhile we state the congruence
condition discussed above as an axiom.
Axiom 1. The Side, Angle, Side (SAS) Congruence Condition. If two sides and
the angle included between these sides are congruent to two sides and the included angle
of the second triangle respectively, then the triangles are congruent.
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Using the SAS condition we can prove other congruence conditions, but first we need
a theorem concerning angles of an isosceles triangle. (A triangle is isosceles if at least
two of its sides are congruent. If all the sides of a triangle are congruent, the triangle is
equilateral. A triangle with no two sides congruent is called scalene. A triangle with
all acute angles is called an acute triangle. A triangle with an obtuse angle is called an
obtuse triangle.)
Theorem 1.3. The Isosceles Triangle Theorem. If two sides of a triangle are congruent, then the angles opposite these sides are congruent.
Restatement: The theorem is frequently stated as follows: The base angles of an
isosceles triangle are congruent.
The markings in Figure 1-11(a) give a quick pictorial content of the theorem. The
marks on the sides indicate the hypothesis that the sides are congruent and the marks
with exclamations on the base angles indicate the conclusion that the angles are congruent.
Whenever appropriate exclamation marks will be used to indicate the conclusion.
(a)
(b)
(c)
Figure 1-11
We formally state the hypothesis and the conclusion of the theorem:
Hypothesis: In "BAC (see Figure 1-11 (a)), AB ∼
= AC
Conclusion: ∠B ∼
= ∠C.
Plan: One approach is to “divide” "ABC into two triangles that seem to be congruent
and in which ∠B and ∠C will be corresponding angles. Anticipating the use of the SAS
condition we bisect ∠A in Figure 1-11 (b). If we could show that the two triangles are
congruent it would follow that the corresponding angles ∠B and ∠C are congruent.
Proof 1. Bisect ∠A. The angle bisector intersects BC at D as shown in Figure 1-11 (b).
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We have now: "ABD ∼
= "ACD by SAS ( AB ∼
= AC, ∠BAD ∼
= ∠CAD and AD ∼
=
∼
AD). Because corresponding parts of the triangle are congruent, ∠ABD = ∠ACD.
To motivate the second approach imagine that "ABC in Figure 1-11 (a) is physically
picked up and turned over. The triangle and the labeling of the vertices would now look
like Figure 1-11 (c). (The same result can be accomplished by tracing "ABC in Figure 111 (a) on the other side of a semi-transparent paper or transparency or by reflecting the
triangle in Figure 1-11 (b) in a mirror placed on AD and perpendicular to the plane ABC.)
The triangles in Figures 1-11 (a) and (c) are congruent and therefore their corresponding
parts are congruent.
Proof 2. "BAC ∼
= "CAB by SAS because BA ∼
= CA, ∠BAC ∼
= ∠CAB and AC ∼
= AB.
∼
Since the corresponding angles in these triangles are congruent we have ∠B = ∠C.
Remark. You may feel uneasy about the second proof of Theorem 1.3; perhaps because
you are used to seeing two separate triangles when you are proving that triangles are
congruent. However, note that the definition of congruence of triangles does not preclude
the possibility that if BA ∼
= CA, "BAC ∼
= "CAB as indicated in the second proof.
Notice that if "BAC is not isosceles then "BAC ! "CAB.
Theorem 1.3 implies the following:
Corollary 1.1. All the angles of an equilateral triangle are congruent, that is an equilateral
triangle is equiangular.
We are now ready to prove two other congruence conditions for triangles.
Theorem 1.4. The Angle Side Angle (ASA) Condition. Given a one-to-one correspondence between the vertices of two triangles, if two angles and the included side of one
triangle are congruent to the corresponding parts of the second triangle, the two triangles
are congruent.
Restatement. Given: "ABC and "A1 B1 C1 and the correspondence A ↔ A1 , B ↔ B1 ,
C ↔ C1 , ∠A ∼
= ∠A1 , AB ∼
= A1 B1 , ∠B ∼
= ∠B1 . Prove: "ABC ∼
= "A1 B1 C1 .
Plan: Because the only congruence condition we can use is the SAS condition we try in
Figure 1-12 (b) to construct an additional side so that SAS can be applied. One way is to
construct a point C2 so that A1 C2 ∼
= AC. Then by SAS, "ABC ∼
= "A1 B1 C2 . We then
prove that C2 = C1 .
−−−→
Proof. Construct on A1 C1 in Figure 1-12 (b) a point C2 so that A1 C2 = AC. By SAS
(regardless if C2 (= C1 or C2 = C1 ) we have:
(1) "ABC ∼
= "A1 B1 C2 .
Pause &
Consider
Why
22
(a)
(b)
Figure 1-12
To show that C2 = C1 notice that (1) implies:
(2) ∠A1 B1 C2 ∼
= ∠ABC.
In addition the hypothesis tells us that:
(3) ∠ABC ∼
= ∠A1 B1 C1 .
Now (2) and (3) with the transitive property of congruence imply ∠A1 B1 C1 ∼
= ∠A1 B1 C2 .
We show now that this congruence of angles implies C2 = C1 . Indeed if C1 (= C2 , then
either C1 is between A1 and C2 or C2 is between A1 and C1 . In the first case C1 would
be in the interior of ∠A1 B1 C2 which would imply that m(∠A1 B1 C1 ) < m(∠A1 B1 C2 ) and
would contradict the fact that the angle are congruent. Similarly we can show that C2
between A1 and C1 contradicts equality of the angles. Thus C2 = C1 . Consequently from
(1) we obtain that "ABC ∼
= "A1 B1 C1 .
NOW PROVE THIS 1-1:
the ASA condition.
Prove the following statements which follow from
(a) Converse of Theorem 1.4: If two angles of a triangle are congruent, then the
sides opposite these angles are congruent.
(b) An equiangular triangle is equilateral (all sides congruent).
23
Medians, Altitudes and Additional Properties of Isosceles Triangles
The segment connecting the vertex of a triangle with the midpoint of the opposite side is
called a median. If M is the midpoint of BC in Figure 1-13 (a), then AM is a median.
The segment from a vertex of a triangle perpendicular to the line containing the opposite
side is an altitude of the triangle. In Figure 1-13 (a) AH⊥BC thus AH is the altitude
to side BC. In Figure 1-13 (b) AH is perpendicular to the line containing BC and hence
is also an altitude to BC. Every triangle has three medians and three altitudes.
(a)
(b)
Figure 1-13
←→
In Figure 1-14 line ! is perpendicular to AB and bisects AB, (AM ∼
= M B). Such a
line is the perpendicular bisector of AB.
Figure 1-14
24
Paper Folding Investigation: To investigate the relationship among a median,
altitude, angle bisector, and perpendicular bisector of a base in an isosceles triangle,
fold or draw an isosceles triangle ABC as in Figure 1-15 (a) where AB = AC and
BC is the base. Fold the paper so that C falls on top of B and then unfold it. The
unfolded triangle and the crease are shown in Figure 1-15 (b).
(a)
(b)
Figure 1-15
While performing the above paper folding activity you have likely noticed that "ACM
was folded onto "ABM (see Figure 1-15 (b)). Thus it seems that ∠CAM ∼
= ∠BAM ,
∼
∼
BM = M C, and ∠AM B = ∠AM C. The last congruence of angles means that AM ⊥BC.
These observations tell us that in an isosceles triangle the angle bisector of the angle
opposite the base is also the perpendicular bisector to the base and the altitude to the
base. We state these observations in the following theorem.
Theorem 1.5. The median to the base of an isosceles triangle is the perpendicular bisector
as well as the angle bisector of the angle opposite the base.
Pause &
Consider
Why
−−→
Proof . Assume that in Figure 1-15 (b) AM is the angle bisector of ∠A. It is sufficient to
prove that "ABM ∼
= "ACM .The congruency of these triangles follows from SAS.
Remark. In the proof of Theorem 1.5 and from now on we assume that an angle has a
unique angle bisector. This fact can be proved (as in Moise p. 109).
25
Theorem 1.5 can be viewed as follows: If the vertex A in Figure 1-15 (b) is equidistant
from the endpoints of the segment BC (that is AC = AB) then A lies on the perpendicular
bisector of BC. It seems that the converse which we state in the next theorem is also true.
Theorem 1.6. Every point on the perpendicular bisector of a segment is equidistant from
the endpoints of the segment.
Restatement. Given: m the perpendicular bisector of AB (see Figure 1-16) and P is
any point on m. Prove: AP ∼
= BP .
Figure 1-16
Proof . There are two cases. If P is on AB then since it is on the perpendicular bisector
of AB, it must be the midpoint of AB and hence equidistant from A and B. If P is not on
AB as in figure 1-16 we have: "AM P ∼
= "BM P by SAS since AM ∼
= BM , M P ∼
= MP,
and the included angles at M are congruent as each is a right angle. From the congruence
of the triangles, we have AP ∼
= BP .
Theorem 1.6 tells us that every point on the perpendicular bisector of a segment is
equidistant from the endpoints of the segment. Are there any other such points? That is,
are there points not on the perpendicular bisector which are equidistant from the endpoints
of the segment? By Theorem 1.5, if P is equidistant from A and B, it must be on the
perpendicular bisector of AB. Consequently, the perpendicular bisector of a segment is
the set of all the points equidistant from the endpoints of the segment. In geometry, a set
of points satisfying a certain property is often called a locus. Thus we have the following:
Corollary 1.2 (to Theorems 1.5 and 1.6). A point is equidistant from the endpoints
of a segment if and only if it is on the perpendicular bisector of the segment. Equivalently
the locus of all the points equidistant from the endpoints of a segment is the perpendicular
bisector of the segment.
26
Figure 1-17 shows two points P and Q equidistant from A and B. From Corollary
1.2 we know that P and Q are on the perpendicular bisector of AB. Because a line is
Figure 1-17
determined by two points, any two points on the perpendicular bisector of a segment
←→
determine the perpendicular bisector. Hence P Q is the perpendicular bisector of AB.
This is stated in the following corollary to Theorems 1.5 and 1.6.
GSP 1-1
Corollary 1.3. If each of two points is equidistant from the endpoints of a segment, then
the line through these points is the perpendicular bisector of the segment.
NOW CONSTRUCT THIS 1-1
(a) Use any tools to construct an acute scalene triangle and the three perpendicular
bisectors, three angle bisectors, three medians, and three altitudes. What do
you notice?
(b) Repeat part (a) for an obtuse triangle.
Construction of a Perpendicular Bisector of a Segment
Corollary 1.3 tells us that if we find any two points equidistant from the endpoints of
a segment the line through the points is the perpendicular bisector of the segment. In
Figure 1-18 two such points P and Q have been constructed using a compass (see the
←→
construction of an isosceles triangle earlier in this section). P Q is the perpendicular
bisector of AB.
27
Figure 1-18
Construction of a Perpendicular to a Line Through a Point on the Line
In Figure 1-19 (a), M is a point on ! and we need to construct the perpendicular to !
through M . Our strategy uses the previous construction. For that purpose we create
an arbitrary segment AB so that M is its midpoint. This is done in Figure 1-19 (b) by
drawing a circle (or semicircle) centered at M . The circle intersects ! at A and B. Thus
M is equidistant from A and B. We now only need to find another point P equidistant
←−→
from A and B. The line PM is the perpendicular bisector of AB and hence as required
perpendicular to ! through M .
(a)
(b)
Figure 1-19
28
Properties of a Kite
When each of two points is equidistant from the endpoints of a segment and none is on
the segment we obtain a special quadrilateral called a kite. It can also be described as
a quadrilateral created by two isosceles triangles sharing a common base. In Figure 1-20
(a), ABCD is a convex kite and in Figure 1-20 (b), ABCD is a non-convex kite.
(a)
(b)
Figure 1-20
If we fold each of the kites in Figure 1-20 along the diagonal AC or the line containing
AC, we observe that B falls on D and hence the diagonals of a kite are perpendicular to
each other, AC bisects BD and bisects the angles at A and C. We state these properties
in the following theorem.
Theorem 1.7. The diagonal of a kite connecting the vertices where the congruent sides
intersect bisects the angles at these vertices and is the perpendicular bisector of the other
diagonal.
Proof. In Figure 1-20(a) or (b) each of the vertices A and C are equidistant from the
←→
endpoints of BD. Hence AC is the perpendicular bisector of BD (Corollary 1.3). Because
the perpendicular bisector of the base of an isosceles triangle is also an angle bisector of
←→
the angle opposite the base, AC bisects each of the angles at A and C. In Figure 1-20 (b),
←→
AC bisects ∠BAD which implies that ∠BAC ∼
= ∠DAC.
A quadrilateral in which all the sides are congruent is called a rhombus. Since a
rhombus is a kite, we have the following corollary.
Corollary 1.4. The diagonals of a rhombus are perpendicular bisectors of each other and
each bisects a pair of opposite angles.
29
NOW PROVE THIS 1-2:
(a) Prove the converse of the statement in Corollary 1.4 A quadrilateral in which
the diagonals are perpendicular bisectors of each other is a rhombus.
(b) State and prove a statement similar to that in part (a) for a kite.
(c) Classify each of the following quadrilaterals using angle bisectors. (Your statement should start like the one in part (a).) (i) Rhombus (ii) Kite Justify your
answers.
Basic Constructions Using Properties of a Kite
The properties of a kite or a rhombus can be used to perform some basic constructions.
To bisect a given segment AB, we construct a kite or a rhombus so that AB is the
common base of the two isosceles triangles as in Figure 1-21 (a). The diagonal DC will
be the perpendicular bisector of AB.
(a)
(b)
Figure 1-21
To bisect a given angle like ∠A in Figure 1-21 (b) we create a kite (or a rhombus)
in which A is the vertex where two congruent sides of a kite intersect. This is done by
constructing any two isosceles triangles with a common base. The diagonal AC will bisect
the angles at A and C. Other constructions utilizing the properties of a kite will be
investigated in the problem set.
30
Remark. Notice that the above constructions utilizing the properties of a kite involve
basically the same steps as the corresponding ones done earlier using the properties of
a perpendicular bisector and an isosceles triangle. Because a kite (or a rhombus) is a
concrete and appealing object some students find it easier to recall how to perform these
constructions by referring to the properties of a kite (or a rhombus).
The properties of a kite can be used to prove the Side-Side-Side congruency condition
stated in the following theorem.
Theorem 1.8. Side-Side-Side (SSS) Congruency Condition. Given a one-to-one
correspondence among the vertices of two triangles, if the three sides of one triangle are
congruent to the corresponding sides of the second triangle, then the triangles are congruent.
Restatement. Given: "ABC, "A1 B1 C1 , and the correspondence A ↔ A1 , B ↔ B1 ,
C ↔ C1 and AB ∼
= A1 B1 , AC ∼
= A1 C1 , BC ∼
= B1 C1 . Prove: "ABC ∼
= "A1 B1 C1 .
(a)
(b)
Figure 1-22
Plan: The idea is to copy "A1 B1 C1 so that the copy and "ABC share a side (side AC
in Figure 1-22). The copy of "A1 B1 C1 is flipped over. In this way a kite is formed and
we can use what we know about kites to show that the triangles are congruent.
Pause &
Consider
Why
Proof. In Figure 1-22 (b) construct triangle III congruent to triangle I using SAS. This
can be accomplished by constructing ∠CAB2 congruent to ∠C1 A1 B1 and AB2 = A1 B1 .
It follows now that ABCB2 is a kite . (AB2 = A1 B1 by construction, A1 B1 = AB (given)
31
and hence AB2 = AB. Also B2 C = B1 C1 (CPCT) and B1 C1 = BC (given), which implies
B2 C = BC). From properties of a kite CA bisects ∠BCB2 and consequently triangles III
and II are congruent by ASA. From the congruence I ∼
= III and III ∼
= II it follows (by
transitivity) that I ∼
= II i.i. that "A1 B1 C1 ∼
= "ABC.
An approach similar to the one used in the proof of the SSS theorem can be used to
determine a useful condition for two right triangles to be congruent. A triangle in which
one of the angles is a 90◦ angle that is a right angle is called a right triangle. The
side opposite the right angle is called the hypotenuse and the other two sides are called
legs. (We will soon prove that a triangle cannot have two right angles.)
Theorem 1.9. Hypotenuse-Leg (H-L) Congruence Condition. If the hypotenuse
and a leg of one triangle are congruent to the hypotenuse and a leg of another right triangle,
then the triangles are congruent. (The proof of this theorem is suggested in Figure 1-23
and will be explored in the problem set.)
(a)
(b)
Figure 1-23
The Relative Size of an Exterior Angle of a Triangle
When a side of a triangle is extended as in Figure 1-24 an angle supplementary to an angle
of a triangle is formed. Such an angle is called an exterior angle of the triangle. ∠BAC is
one of the angles of the triangle and ∠BAD as well as ∠CAE are exterior angles. Notice
that ∠BAD and ∠CAE are vertical angles and hence congruent. In general a triangle has
three noncongruent exterior angles.
Any angle of the triangle that is not adjacent to an exterior angle is called a remote
interior angle of the exterior angle. Thus ∠B and ∠C are remote interior angles of
32
Figure 1-24
∠BAD. The following theorem compares the size of an exterior angle and a remote
interior angle. (We say that one angle is greater than another if the measure of that angle
is greater than the measure of the other.) The following theorem compares the size of an
exterior angle and its remote interior angle. It is a major theorem as it will be used to
prove several other theorems.
Theorem 1.10. The Exterior Angle Theorem. An exterior angle of a triangle is
greater than either of the remote interior angles.
Restatement. Given: "ABC in which BA has been extended as in Figure 1-25. Prove:
m(∠CAD) > m(∠ACB).
Plan: To prove that ∠CAD is greater than ∠C we try to “fit” the latter angle inside the
exterior angle. This can be done by creating congruent triangles.
Figure 1-25
∼ MN.
Proof. Let M be the midpoint of AC. Extend BM at its length, so that BM =
∼
∼
Now "M CB = "M AN by SAS (why?) and hence ∠CAN = ∠C. Because N is in the
interior of ∠CAD, ∠CAN is smaller than ∠CAD and since ∠CAN ∼
= ∠C, it follows that
33
∠C is smaller than ∠CAD. To prove that ∠CAD is greater than ∠B extend CA to create
−→
the ray CA. Then an exterior angle congruent to ∠CAD is formed. Now “fit” ∠B inside
that angle by a proper construction. The details are left as an exercise.
Notice that we have not justified the intuitively obvious fact that N is in the interior of
∠CAD. This can be proved using The Plane Separation Axiom of Appendix I. We leave
the details to the interested reader.
The following corollaries follow immediately from the Exterior Angle Theorem.
Corollary 1.5. Through a point not on a line there is a unique (one and only one)
perpendicular to the line.
(a)
(b)
Figure 1-26
Restatement. Given: a line ! as in Figure 1-26 (a) and P a point not on !. Prove:
there exists one and only one line through P perpendicular to !.
Proof. The existence of a line through P perpendicular to ! can be established by constructing a segment on ! so that P is equidistant from the endpoints of the segment and
then constructing another point equidistant from the same endpoints. (This is explored
in Now Construct This 1-2.)
NOW CONSTRUCT THIS 1-2
In Figure 1-26 construct a circle with center at P intersecting ! in two points. Next
←→
construct a point Q equidistant from these two points. Why is P Q perpendicular to
!?
34
To prove the uniqueness assume that there is more than one perpendicular through
P . Let P A and P B be two perpendiculars to ! as shown in Figure 1-26 (b). Then in
"P AC the exterior angle P BC and its remote interior angle P BA are right angles. This
contradicts the Exterior Angle Theorem. Consequently more than one perpendicular to !
through P cannot exist.
The Exterior Angle Theorem or the uniqueness of the perpendicular to a line through
a point can be used to prove the following:
Theorem 1.11. Hypotenuse-Acute Angle Congruence Condition: If the hypotenuseand an acute angle of one right triangle are congruent to the hypothenuse and an acute
angle of another right triangle, then the triangles are congruent.
Figure 1-27
Restatement. Given: "ABC and "A1 B1 C1 in which AB ∼
= A1 B1 , ∠A ∼
= ∠A1 , ∠C
and ∠C1 are right angles. Prove: AC ∼
= A1 C1 .
Proof. In Figure 1-27, the marked parts of "ABC and "A1 B1 C1 are congruent. If
AC ∼
= A1 C1 , then the triangles are congruent by SAS. If AC (= A1 C1 , then AC < A1 C1 or
−→
AC > A1 C1 . In either case we can find a point C2 on AC such that AC2 ∼
= A1 C1 . Then
"ABC2 ∼
= "A1 B1 C1 by SAS and hence ∠AC2 B is a right angle. Consequently BC2 and
←→
BC are two different perpendiculars from B to AC. This contradicts the uniqueness of a
perpendicular from a point to a line. Thus AC (= A1 C1 must be rejected and therefore
AC ∼
= A1 C1 and "ABC ∼
= "A1 B1 C1 by SAS.
The existence of a unique line perpendicular to a given line through a given point
enables us to define a concept important in finding areas of triangles and other figures.
35
Definition of the Distance From a Point to a Line
The distance from a point P to a line ! is the length of the segment connecting P with the foot of the perpendicular to ! through P . (The foot of the
perpendicular is the point of intersection of the perpendicular and !.)
In Figure 1-28, the length of P Q is the distance from P to !.
Figure 1-28
The distance from a vertex of a triangle to the line containing the opposite side is
called a height. Thus a height is the length of an altitude. A triangle has three heights
corresponding to its three sides. Figure 1-29 shows the three altitudes of an obtuse triangle
and the corresponding heights h1 , h2 , and h3 . Notice that the lines containing the three
Figure 1-29
altitudes intersect at a single point P in Figure 1-29. When three or more lines intersect
in a single point we say that they are concurrent. The proof of the fact that the lines
containing the altitudes of a triangle are concurrent follows from the fact that the perpendicular bisectors of the sides of a triangle are concurrent. These and other concurrency
theorems will be explored in the problem set.
GSP 1-2
36
Paper Folding Investigation
(a) Draw a large acute scalene triangle on a sheet of semi-transparent paper and fold
it to create the three perpendicular bisectors of the sides. Repeat the exercise
for an obtuse triangle. What do you notice?
(b) Repeat the experiment in part (a) for altitudes of an acute and an obtuse triangle.
(c) Repeat the experiment in part (a) for the three angle bisectors.
(d) Draw any angle and crease the paper to create the angle bisector. Mark a point
on the angle bisector and crease the paper to create a segment whose length is
the distance from the point to the sides of the angle. Fold the paper again along
the angle bisector so that the sides of the angle coincide. What do you notice
about the distance from the point to the sides of the angle?
We have seen earlier that a perpendicular bisector of a segment can be characterized as
follows: A point is on the perpendicular bisector of a segment if and only if it is equidistant
from the endpoints of the segment. Can an angle bisector be characterized in a similar
way? The definition of the distance between a point and a line enables us to do so as seen
in the following theorem.
Theorem 1.12. A point is on the angle bisector of an angle if and only if it is equidistant
from the sides of the angle.
Figure 1-30
Restatement.
(i) Given: P on the angle bisector of ∠A as in Figure 1-30. Prove: P B ∼
= P C (P B
and P C are perpendiculars to the sides of the angle respectively).
37
(ii) And conversely: Given: the fact that the distances from P to the sides of the angle
are equal. Prove: P is on the angle bisector of ∠A.
Proof of (i). Because P is on the angle bisector of ∠A, "AP B ∼
= "ACP by HypothenuseAcute Angle Theorem. Consequently, P B = P C (by CPCT).
Proof of (ii) Since P B ∼
= overlineP C (given), "AP B ∼
= "ACP by Hypothenuse-Leg
Condition. Consequently ∠P AB ∼
∠P
AC
(by
CPCT).
=
Using the Exterior Angle Theorem we can prove a theorem concerning the relative size
of the angles opposite non-congruent sides of a triangle and conversely the relative size of
the sides opposite noncongruent angles.The proofs of these theorems will be explored in
the problem set.
Theorem 1.13. Given two non-congruent sides in a triangle, the angle opposite the longer
side is greater than the angle opposite the shorter side.
Theorem 1.14 (The converse of Theorem 1.13). Given two non-congruent angles in
a triangle, the side opposite the greater angle is longer than the side opposite the smaller
angle.
The last theorem is instrumental in proving the intuitively well known fact that the
shortest path connecting any two points A and B is the segment AB. This is formalized
in the following theorem which is fundamental in most branches of mathematics.
Theorem 1.15 (The Triangle Inequality). The sum of the length of any two sides of
a triangle is greater than the length of the third side.
(a)
(b)
Figure 1-31
38
Restatement. Given: "ABC with sides of length a, b, and c (shown in Figure 1-31
(a)). Prove: a + b > c, a + c > b, and b + c > a.
Plan: We prove the first inequality a + b > c. (The proof of the other two inequalities
are analogous.) Our plan is to construct a triangle with one side a + b and another side
c. Then to show that the angle opposite side a + b is larger than the angle opposite side
c and hence conclude from Theorem 1.14 that a + b > c.
Proof. Extend side BC in Figure 1-31 (b) so that CD = b. Next we show that ∠BAD
is greater than ∠D. We know that ∠D ∼
= ∠CAD since "CDA is isosceles. Because C is
in the interior of ∠BAD, we have: m(∠BAD) > m(∠CAD). Consequently m(∠BAD) >
m(∠D). Applying Theorem 1.14 in "BAD we have BD > BA or a + b > c.
Example 1.2. Given line ! and points A and B not on the line, find the shortest path
from A to a point on the line and then to B. (See also Chapter 0 Problem 5 - The Hikers
Path).
(a)
(b)
Figure 1-32
Solution.
Case I. In Figure 1-32 (a) the points A and B are on opposite sides of the line. The
segment AB intersects ! at P and the “straight” path A − P − B is the shortest path.
This is the case because if Q is any other point on !, then by the triangle inequality:
AQ + QB > AB.
Case II. Suppose that A and B are on the same side of ! as in Figure 1-32 (b). We
could reduce this case to the previous one by imagining that ! is a mirror and finding the
shortest path connecting A with a point P on ! and the reflection B " of B in the mirror.
The reflection of B in ! is obtained by dropping a perpendicular from B to ! and finding
39
B " so that ! is the perpendicular bisector of BB " . (A detailed study of reflections will be
pursued in a later chapter on Transformations.)
Notice that a property of a perpendicular bisector implies that if Q is any point on !
then the length of the path A − Q − B is the same as that of A − Q − B " . Consequently
this case is reduced to finding the shortest path connecting A, to a point Q on ! and B " .
This can be solved using Case I by connecting A with B " and finding the point P where
AB " intersects !. The path A − P − B is the required path.
In the solution of Example 1.2 we used the concept of reflection in a line. This concept
will be used again in Chapter 5 on transformations.
Definition of reflection in a Line
A reflection in a line ! assigns to each point P in the plane the point P " , the
image of P , in such a way that ! is the perpendicular bisector of P P " as long
as P is not on the line. If P is on ! then P " = P (See Figure 1-33)
(a)
(b)
Figure 1-33
Remark. (i) When P " is the image of P under reflection in a line ! we also say that
P " is the reflection of P in !.
(ii) Notice that reflection in a line is a function from the plane to the plane.
Pause &
Consider
Why
40
Properties of Parallel Lines
Through a point P not on a line ! there exists a line parallel to !. This fact follows from
the following theorem.
Theorem 1.16. If each of two lines in the same plane is perpendicular to a third line in
that plane, then they are parallel.
Figure 1-34
Restatement. Given: k, !, and m three lines in the same plane as in Figure 1-34 such
that k⊥m and !⊥m. Prove: k||!.
Proof. If k and ! are not parallel then they must intersect at some point C. In that
case, there will be two perpendiculars to m through C, which contradicts Corollary 1.5.
Consequently k||!.
Theorem 1.16 can also be proved by direct use of the Exterior Angle Theorem. The
angles marked in Figure 1-34 are 90◦ each and hence ∠CAB is also a right angle (property
of vertical angles). Consequently an exterior angle of "ABC is equal to a remote interior
angle, which contradicts the Exterior Angle Theorem. Notice that if the marked angles
in Figure 1-34 were only congruent but not necessarily 90◦ each, we could use the same
argument to prove k||!. This is stated in the next theorem but before we proceed we need
to define some commonly used terminology for angles formed when a line intersects two
given lines.
Any line that intersects a pair of lines in two points is called a transversal of those
lines. In Figure 1-35, line m is a transversal of lines k and !. Angles are formed by these
lines and are named according to their position relative to the transversal.
41
Figure 1-35
Interior Angles: ∠1, ∠3, ∠6, ∠8
Corresponding Angles: ∠1 and ∠2, ∠3 and ∠4,
∠5 and ∠6, ∠7 and ∠8
Alternate Interior Angles: ∠1 and ∠8, ∠3 and ∠6
Alternate Exterior Angles: ∠5 and ∠4, ∠7 and ∠2
Remark. Notice that a pair of corresponding angles are congruent if and only if a pair
of alternate interior angles are congruent.
Theorem 1.17. If two lines are cut by a transversal and a pair of corresponding angles
are congruent (or a pair of alternate interior angles are congruent) then the lines are
parallel.
The proof of Theorem 1.17 is explored next in Now Prove This 1-3.
NOW PROVE THIS 1-3: Prove Theorem 1.17 by contradiction, showing that if
the lines were not parallel, a triangle would be formed with an exterior angle congruent
to a remote interior angle which will contradict the Exterior Angle Theorem.
The reader may wonder at the converse of theorem 1.16: If two parallel lines are cut
by a transversal then a pair of corresponding angles are congruent.
This statement is needed to prove that the sum of measures of the angles of any
triangle is 180◦ and consequently many other theorems of Euclidean geometry. In fact
the statement is equivalent to the famous Parallel Postulate discussed in the next section
and hence distinguishes Euclidean geometry from non-Euclidean geometries. (See the
Historical Note below.) Even the existence of rectangles cannot be established using the
axioms we have stated and theorems proved so far. (A rectangle is a quadrilateral with
42
four right angles.) If we try to construct a rectangle as in Figure 1-36 by constructing a
line !, marking two points A and B on the line and then constructing two segments AD
and BC each perpendicular to !, we obtain what is called a Saccheri quadrilateral.
The segment AB is called the lower base (at A and B the angles are 90◦ ) and the opposite
side is called the upper base.
Figure 1-36
It is easy to prove that the diagonals of a Saccheri quadrilateral are congruent and
that the upper base angles are congruent. Because we have not proved yet that the sum
of the measures of the angles of a triangle is 180◦ (and consequently that the sum of the
measures in any quadrilateral is 360◦ ), we cannot conclude that the Saccheri quadrilateral
is a rectangle. This can be done with the Parallel Postulate, an axiom equivalent to
Euclid’s Fifth Postulate.
Historical Note: Gerolamo Saccheri
Gerolamo Saccheri (1667-1733), an Italian mathematician, was dissatisfied with the
parallel postulate. He, like most mathematician at the time, believed that it should
be possible to prove the postulate as a theorem using previous axioms and theorems.
He wrote a book entitled Euclid Freed of All Blemish in which he tried to prove the
Parallel Postulate. Saccheri’s proof was wrong but in the attempt to find a proof he
did prove many new theorems which are now considered a part of absolute or neutral
geometry, a geometry independent of the parallel postulate. The theorems we have
proved so far belong to the realm of absolute or neutral geometry.
Problem Set 1.2
Problems 1 through 19 should be answered without the use of the Parallel Postulate.
∼ "DEF can also be written "BAC =
∼ "EDF .
1. Notice that the congruence "ABC =
How many such symbolic representations are there between two congruent triangles?
Do different representations give different information about the triangles?
43
2. State and prove a congruence condition (other than the definition) for two quadrilaterals to be congruent.
3. State and prove a theorem analogous to Theorem 1.7 for non-convex kites.
4. Write a careful proof of Theorem 1.8, (SSS condition).
5. Write a proof of Theorem 1.9, (Hypothenuse-Leg condition).
6. (a) Draw an acute triangle (all angles less than 90◦ ) and an obtuse triangle (one
angle more than 90◦ ) and in each case construct the three perpendicular bisector
of the sides of the triangle.
(b) In part (a) you must have noticed that the perpendicular bisectors of the sides
of a triangle are concurrent, that is intersect in a single point. Examine the
following proof that this is the case and use a similar approach o prove that the
angle bisectors of a triangle are concurrent.
In "ABC let O be the point where the perpendicular bisectors of AB
and BC intersect. We only need to prove that O is also on the third
perpendicular. We have proved that a point is on the perpendicular
bisector of a segment if and only if it is equidistant from the endpoints
of the segment. Thus it suffices to prove that AO = CO. We have:
AO = BO (since O is on the perpendicular bisector of AB)
BO = CO (since O is on the perpendicular bisector of BC)
Hence AO = CO and therefore O must be on the perpendicular bisector of AC.
Figure 1-37
(c) The proof in part (b) is not completely rigorous; an assumption was made
without proving it. What was the assumption?
GSP 1-3
44
7. A circle that passes through each vertex of a triangle is called a circumscribing
circle as shown in Figure 1-38 (a). A circle that is tangent to each side of a triangle
is an inscribed circle as shown in Figure 1-38 (b).
(a) Assume that a tangent to a circle is perpendicular to the radius at the point
of contact and explain how to find the circumscribing circle and the inscribed
circle for a given triangle. Justify your answers.
GSP 1-4
(b) Draw a triangle in which all the angles seem to be acute (less than 90◦ ) and a
triangle in which one angle is obtuse (greater than 90◦ ). In each case construct
the circumscribing and inscribed circles.
(a)
(b)
Figure 1-38
8. ABCD is a parallelogram. Its diagonals divide the parallelogram into 4 triangles.
Points F , G, H, I are the incenters (the centers of the inscribed circles) of the
corresponding triangles. What is the most that can be concluded about the type of
quadrilateral F GHI is? Prove your answer.
Figure 1-39
45
9. Use the Exterior Angle Theorem to prove that in any "ABC : m(∠A) + m(∠B) +
m(∠C) < 270◦ .
∗∗ 10.
Without using the Parallel Postulate it is possible to prove that in any "ABC,
m(∠A) + m(∠B) + m(∠C) ≤ 180◦ . This can be shown by proving first that in any
Saccheri quadrilateral, the upper base is as long or longer than the lower base. Prove
these assertions by referring to one of the texts listed in the bibliography. (The text
by Moise is a good choice.)
10. Prove that in any quadrilateral the sum of the lengths of any three sides is greater
than the length of the fourth side.
11. Refer to diagram (i) in problem 11 to prove Theorem 1.14 (which is the converse of
Theorem 1.13) that α > β ⇒ a > b. Assume that a > b is false, i.e., that a ≤ b and
show that the case a = b as well as a < b cause a contradiction (the latter contradicts
Theorem 1.13.)
12. Explain how the proof of Theorem 1.14 can be accomplished by proving the contrapositive (see Appendix 2) of the statement α > β ⇒ a > b.
1.3
The Parallel Postulate and Its Consequences
Circa 300 B.C. Euclid arranged plane geometry based on five postulates. The fifth postulate known as the parallel postulate was stated by Euclid as follows:
If a straight line falling on two straight lines make the interior angles on the
same side less than two right angle, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.
From ancient times this postulate was criticized, Euclid proved 28 theorems before using
the fifth postulate in a proof. The converse of the parallel postulate, “The sum of two
angles of a triangle is less than two right angles,” was proved in Euclid’s Elements as
a theorem. Consequently for generations mathematicians felt that the parallel postulate
could be proved based on Euclid’s previous postulates. As a result of the many attempts
to prove the parallel postulate several equivalent statements have been discovered. The
most common substitute for Euclid’s Parallel Postulate is known as Playfair’s axiom
after the Scottish mathematician John Playfair (1748-1819), although it appears earlier
in the writing of Proclus (A.D. 410-485) in his Commentary on the First Book of Euclid’s Elemenets. Because of its simplicity Playfair’s axiom is used today as the Parallel
Postulate:
Axiom 2. The Parallel Postulate(Playfair’s Axiom): Given a line and a point not on
the line, there exists a unique line through the point parallel to the given line.
46
Restatement: Given !1 and P not on !1 there exists one and only one line !2 through F
such that !1 ||!2 (see Figure 1-40).
Figure 1-40
Historical Note: John Playfair
John Playfair(1748-1819), a Scottish mathematician and physicist, was educated at
home until the age of 14 when he started his studies at the University of St. Andrews.
He was ordained as a minister but continued his studies and in 1785 was awarded the
chair of mathematics at the University of Edinburg. In 1795 he wrote the Elements of
Geometry which became a standard text in geometry and went through many editions.
In that text, Playfair introduced his version of parallel postulate which is standard
today.
The Parallel Postulate is the turning point at which different geometries branch out.
By accepting the Parallel Postulate, Euclidean Geometry is developed. If we deny the
postulate there are two possibilities:
(1) There is more than one parallel to a line through a given point not on the line
(2) There is no parallel to a line through a point not on the line.
In each case we obtain a Non-Euclidean Geometry. Statement (1) is the basis for
Hyperbolic Geometry while statement (2) with some modifications in previous axioms
results in Elliptic Geometry.
Using the Parallel Postulate, the following converse of Theorem 1.17 can be proved
and consequently the subsequent theorems of Euclidean geometry can also be proved.
Theorem 1.18. If two parallel lines are cut by a transversal, then a pair of corresponding
angles are congruent.
Restatement. Given: !1 ||!2 (as in Figure 1-41) and t a transversal. Prove: ∠1 ∼
= ∠2..
Plan: We assume that the angles are not congruent and obtain a contradiction of the
Parallel Postulate.
Proof. If ∠1 and ∠2 are not congruent then there exists a line !3 through P such that
!3 (= !1 and so that ∠4 that !3 makes with t is congruent to ∠1. Then by Theorem 1.17,
47
Figure 1-41
!3 ||!1 , and consequently through P there are two lines !2 and !3 parallel to !1 , which
contradicts the Parallel Postulate. Consequently ∠1 ∼
= ∠2.
Theorem 1.18, its converse Theorem 1.17, and the fact that a pair of corresponding
angles are congruent if and only if a pair of alternate interior angles are congruent imply
each of the following:
Theorem 1.19. Two lines in a plane are parallel if and only if a pair of corresponding
angles formed by a transversal are congruent.
Theorem 1.20. Two lines in a plane are parallel if and only if a pair of alternate interior
angles formed by a transversal are congruent.
In Figure 1-42 !1 ||!2 and the angles are as marked. By Theorem 1.19, α = γ. Because
β + γ = 180◦ we have α + β = 180◦ . Conversely if α + β = 180◦ (and it is not given that
!1 ||!2 ) we can show that α = γ and hence by Theorem 1.19 that !1 ||!2 . These observations
are stated in the following theorem.
Figure 1-42
Pause &
Consider
Why
48
Theorem 1.21. Two lines are parallel if and only if a pair of interior angles on the same
side of a transversal are supplementary.
Restatement: In Figure 1-42, !1 ||!2 if and only if α + β = 180◦ .
We can now prove a key theorem of Euclidean Geometry:
Theorem 1.22. The sum of the measures of the interior angles of a triangle is 180◦ .
Restatement. Given: "ABC. Prove: m(∠A) + m(∠B) + m(∠C) = 180◦ .
Figure 1-43
Proof. We could show the result by showing that the sum of the measures of the angles of a
triangle equals the measure of a straight angle. A straight line will create angles congruent
to the angles of a triangle if it is parallel to one of the sides. Thus we construct in Figure 1←→
43 a line ! through C parallel to AB. Because AC is a transversal for the parallel lines ! and
←→
←→
AB, α = α1 , as they are alternate interior angles. Similarly because CB is a transversal
for the same parallel lines β = β1 . Consequently: α + β + γ = α1 + β1 + γ = 180◦ .
−−→
−→
Example 1.3. In Figure 1-44 !1 ||!2 and k is a transversal. If BC and AC are angle
bisectors of the interior angles as shown, find m(∠B).
Solution. If we mark the measures of the angles as shown in Figure 1-44 we have:
2x + 2y = 180◦
x + y = 90
Now in "ABC:
(Theorem 1.21),
◦
x + y + m(∠B) = 180◦ . Hence
m(∠B) = 180◦ − (x + y)
= 180◦ − 90◦
= 90◦
Consequently ∠B is a right angle.
49
Figure 1-44
NOW PROVE THIS 1-4:
i. Conjecture n expression for the sum of the measures of the interior angles of a
convex n-gon in terms of n.
ii. What is the sum of the measures of the exterior angles of a convex n-gon? Prove
your answer.
We are now able to prove the following:
Theorem 1.23. A Saccheri quadrilateral is a rectangle.
Restatement:
In Figure 1-45 ∠A and ∠B are right angles, and AD ∼
= BC. Show that ∠D and ∠C
are right angles.
Figure 1-45
50
Proof. Construct the diagonal AC. To prove that ∠D is a right angle we show that the
diagonal divides BCD into congruent triangles. Notice that x = y as alternate interior
angles formed by the parallels AD and BC. Hence "ABC ∼
= "CDA by SAS. Consequently ∠D ∼
= ∠B and ∠D is a right angle. Because the sum of the measures of the angles
in any quadrilateral is 360◦ it follows that m(∠C) = 90◦ .
We define now some common terms of Euclidean geometry. A rectangle is one of several
quadrilaterals whose properties will be investigated. Other quadrilaterals include: the
trapezoid, a quadrilateral with at least one pair of parallel sides; the parallelogram, a
quadrilateral in which each pair of opposite sides is parallel; the rhombus: a parallelogram
with two adjacent sides congruent; and the square, a rhombus with a right angle. A
rectangle can be defined as a parallelogram with a right angle. (It is easy to show that
a quadrilateral in which all the angles are right angles is a rectangle, a property used in
Theorem 1.23.)
The proof of the following theorems are straightforward. Readers should prove them
on their own.
Theorem 1.24. The measure of an exterior angle in a triangle is equal to the sum of the
measures of its two remote angles.
Theorem 1.25. If a transversal is perpendicular to one of two parallel lines, it is also
perpendicular to the other line.
Theorem 1.26. In a parallelogram:
(i) Each diagonal divides it into two congruent triangles.
(ii) Each pair of opposite sides are congruent.
(iii) The diagonals bisect each other.
Pause &
Consider
Why
What properties of a parallelogram characterize it? In other words, given a quadrilateral, what conditions will assure that it is a parallelogram? (Of course we should not list
more properties than are necessary.) Before reading on, list as many conditions which are
necessary and sufficient for a quadrilateral to be a parallelogram. The following are some
of the useful conditions. Proofs are left as an exercise.
Theorem 1.27.
(1) A quadrilateral in which each pair of opposite sides are congruent is a parallelogram.
(2) A quadrilateral in which the diagonals bisect each other is a parallelogram.
(3) A quadrilateral in which each pair of opposite angles is congruent is a parallelogram.
51
(4) A quadrilateral in which a pair of opposite sides is parallel and congruent is a parallelogram.
The last condition in Theorem 1.27 may be the least well known but it is often most
useful. Property (1) can be used for an efficient construction of parallel lines. This is
investigated next.
52
NOW PROVE THIS 1-5:
i. Prove Theorems 1.24 through 1.27.
ii. Part (1) of Theorem 1.27 can be written as: A necessary and sufficient condition
for a quadrilateral to be a parallelogram is that each pair of opposite sides are
congruent. Write the other parts of Theorem 1.27 using the phrase “necessary
and sufficient“.
Constructing a line through P (not on !) parallel to !
In Figure 1-46 (a), a line ! and a point P are given. To construct through P a line parallel
to ! we make P a vertex of any parallelogram so that one of the sides of the parallelogram
is on !. By property (1) of Theorem 1.27, to achieve this we need only make each pair
of opposite sides congruent. The simplest is to construct all sides of the quadrilateral
congruent as in Figure 1-46 (b). (That is to construct a rhombus.)
(a)
(b)
(c)
Figure 1-46
To accomplish the construction (see Figure 1-46(c)) we draw any line through P that
intersects ! and label the point of intersection A. (The labeling is of course not necessary
for the actual construction.) We make P and A the first two vertices of the rhombus. We
locate the next vertex B on ! so that AB ∼
= AP . We now find the fourth vertex C so that
PC ∼
= AB ∼
= AP . ( For that purpose we use a compass to construct an arc with center B
and radius AB and an arc with center P and radius AB. The intersection of the arcs is
the point C. Line PC is the required line.)
53
Parallel Projections
If P is a point not on a line ! as in Figure 1-47 (a), then the point P " where the perpendicular through P to ! intersects !, is called the vertical projection of P onto !.
Instead of projecting in the perpendicular direction, we could follow any given direction.
In Figure 1-47 (b) k and ! are two lines and m is a transversal intersecting the lines at Q
and Q" respectively. The line through P parallel to QQ" intersects ! at P " . The point P "
is the parallel projection of P on ! parallel to QQ" . (The parallel projection of Q on
! is Q" .)
(a)
(b)
Figure 1-47
If we consider lines as a set of points, a parallel projection is a function that assigns
to each point of k a corresponding point of !. If we denote the function by f we have
f (P ) = P " and f (Q) = Q" . Notice that the domain of this function is k and the codomain
is !. We could denote this (see Appendix III on functions) by:
f : k −→ !
The function is called the projection of k on ! in the direction of m. In short we refer to
such functions as parallel projections. It is straightforward to show that a parallel projection function is one-to-one and onto hence a one-to-one correspondence. The following
two theorems describe two useful properties of parallel projections. (See Appendix I for
definition of betweeness.)
Theorem 1.28. A parallel projection preserves betweenness.
Restatement. Given: f : k → ! a parallel projection, A, B, C on k, f (A) = A" ,
f (B) = B " , f (C) = C " , and A − B − C (i.e. B is between A and C). Prove: A" − B " − C "
(i.e. B " is between A" and C " ).
This is illustrated in Figure 1-48. Because the result is so plausible we omit the proof.
54
Figure 1-48
Theorem 1.29. A parallel projection preserves congruence of segments.
Restatement. Given: In Figure 1-49 lines k, l and m, A, B, C and D on k so that:
←→ ←−→ ←−→ ←−→
AB = CD. Also A" , B " , C " and D" on ! so that AA" , BB " , CC " , DD" are parallel to m.
(Note: It is not given that AB = BC). Prove: A" B " = C " D" .
Figure 1-49
Proof. We distinguish two cases:
Case 1: If k||! then AB and A" B " are opposite sides of a parallellogram and hence AB = A" B " .
Similarly CD = C " D"" . Since AB = CD, it follows that A" B " = C " D" .
Case 2: If k is not parallel to !, to prove that A" B " = C " D" , we try to construct congruent
triangles with corresponding sides AB and CD as well as sides congruent to A" B "
and C " D" . Keeping in mind the above proved Case 1 this can be accomplished by
constructing through A and C lines p and q parallel to ! as shown in Figure 1-49.
From case 1 it follows that A" B " = AE and C " D" = CF . Hence it will suffice to
55
prove that AE = CF . For that purpose we prove that "ABE ∼
= "CDF . Because each pair of similarly marked angles are corresponding angles between parallel
lines, it follows that ∠ABE ∼
= ∠DCF and hence by ASA that "ABE ∼
= "CDF .
Consequently AE = CF and hence A" B " = C " D" .
In Chapter 4 we will prove that parallel projections also preserve the ratio of the
lengths of the segments as well. Functions preserving certain geometric properties will
play a crucial role in later chapters.
An immediate and useful consequence of the last theorem is obtained when the congruent segments are adjacent, i.e., when B = C in Figure 1-49. We obtain the following:
Corollary 1.6. If three or more parallel lines intercept congruent segments on one transversal, then they intercept congruent segments on any other transversal.
Restatement. If in Figure 1-50, A1 A2 = A2 A3 = A3 A4 = · · · = An−1 An and
A1 A"1 + A2 A"2 + A3 A"3 + . . . + An A"n , then A"1 A"2 = A"2 A"3 = A"3 A"4 = · · · = A"n−1 A"n .
Figure 1-50
Corollary 1.6 can be used for the following construction.
Division of a segment into any number n of congruent parts
Given AB as in Figure 1-51, we illustrate the construction for n = 3.
Figure 1-51
Pause &
Consider
Why
56
−→
−→
Through A construct any ray AC and mark on AC any point P . Construct Q and R
so that AP ∼
= PQ ∼
= QR. Connect R with B and through Q and P draw lines parallel to
BR. The points of intersection P " and Q" accomplish the required construction since by
Corollary 1.6 AP " ∼
= P " Q" ∼
= Q" R.
A special case of Corollary 1.6 for a triangle is:
Corollary 1.7. A line through the midpoint of one side of a triangle and parallel to the
second side bisects the third side.
Restatement. Given: In "ABC, M the midpoint of AC and a line through M parallel
to AB that intersects BC at N as in Figure 1-52. Prove: N is the midpoint of BC.
(We use the double arrow on the segments AB and M N to indicate that the segments are
parallel.).
Figure 1-52
Pause &
Conjecture
The segment connecting the midpoints of two sides of a triangle is called the midsegment. Each triangle has three midsegments. Since a segment has a unique midpoint it
follows from the last corollary that a midsegment of two sides of a triangle is parallel to
the third side. How does the length of the midsegment in a triangle compare to the length
of the parallel side? Since any triangle is “half” of a parallelogram (a diagonal divides a
parallelogram into two congruent triangles) we could complete "ABC in Figure 1-53 (a)
into a parallelogram by tracing "A1 B1 C1 over "ABC and then turning it upside down as
shown in Figure 1-53 (b). (We will see later that this amounts to rotating "ABC about
N by 180◦ .) We obtain a parallelogram if we make B1 = C and C1 = B. It seems now
that M N ∼
= 1/2M M1 and M M1 ∼
= AB and consequently that M N ∼
= 1/2AB. We state
this result in the following theorem along with a more rigorous argument following the
above idea.
Theorem 1.30. The Midsegment Theorem. The segment connecting the midpoints
of two sides of a triangle is parallel to the third side and half as long as that side.
Restatement. Given: "ABC in Figure 1-53 (a), M and N are midpoints of AC and
BC. Prove: (1) M N +AB (2)M N = 1/2 AB.
57
(a)
(b)
Figure 1-53
Proof. Part (1) follows from Corollary 1.7. To prove part (2), we want to obtain a
parallelogram and therefore draw through C a line parallel to AB and through B a line
parallel to AC as shown in Figure 1-53 (b). The lines intersect at A1 . From the definition
it follows that ABA1 C is a parallelogram. We extend M N until it intersects BA1 at M1 .
At this point it seems obvious that N M1 = M N and that M M1 = AB. Thus if M N = x
then 2x = M M1 = AB and hence x = 1/2 AB.
We can prove what we stated as obvious as follows. Theorem 1.29 implies A1 M1 =
C1 M1 and hence N M1 is a midsegment in "A1 BC. We can prove that M N = M1 N by
showing that "M CN ∼
= "M1 BN . Also M M1 = AB since AM M1 C1 is a parallelogram.
Remark. We could have postponed the Midsegment Theorem until chapter 4 and deduced
it from theorems concerning similar triangles but we chose to prove the theorem here as
the methods used give us valuable practice and an opportunity to see different approaches.
NOW PROVE THIS 1-6: We have seen that the diagonals of a parallelogram
bisect each other. Suppose they bisect each other at O.
i. If through O a line is drawn and it intersects two parallel sides of the parallelogram at points P and Q respectively, prove that OP = OQ.
ii. The statement in part (i) is a generalization of a fact stated in the proof of
Theorem 1.29. What fact is it? Why part (i) is a generalization of that fact?
Pause &
Consider
Why
58
Medians of a Triangle
GSP 1-5
A median of a triangle is a segment connecting a vertex to the midpoint of the opposite
side. The Midsegment Theorem can be used to prove a surprising property of the medians
of a triangle. If you construct the three medians of an arbitrary triangle you will notice that
the medians are concurrent. In search for the possibility of other properties of medians we
construct two medians AM and CN in "ABC as shown in Figure 1-54. As in Figure 1-53
(b) we trace "ACB to obtain "A1 C1 B1 place it “upside down” so that B1 = C and
C1 = B.
Figure 1-54
The corresponding medians in "A1 BC are A1 M and BN1 . We obtain the parallelogram ABA1 C and because the diagonals of a parallelogram bisect each other A, M ,
and A1 are collinear. Notice that CN1 BN is also a parallelogram as CN1 ∼
= N B and
CN1 +N B1 (part 3 of Theorem 1.27). Because CN1 ∼
= N1 A1 and AN ∼
= N B, the parallels
CN and N1 B mark congruent segments on OA1 and on AO1 . Thus:
(1) OO1 ∼
= O1 A1
(2) AO ∼
= OO1 .
(3) Consequently: AO ∼
= OO1 ∼
= O1 A1
Because AA1 = 2AM , from (3) we have: AO ∼
= 1/3AA1 = 1/3 · 2 AM = 2/3 AM . Hence
OM = 1/3AM . (Equivalently from Now Prove This 1-6 we could argue that because
OM = M O1 , OM = 1/2 OO1 = 1/2 AO.)
The above investigation suggests the following theorem and its proof. (Two different
proofs will be investigated in the problem set.)
Theorem 1.31. Property of Medians. The medians of a triangle are concurrent in a
point whose distance to a vertex is two-thirds of the length of the median from that vertex.
Restatement: If AM , CN , and BP are the three medians of "ABC as shown in Figure 155 (a), then they intersect in a single point O such that: AO = 2/3AM , CO = 2/3CN ,
59
Pause &
Consider
Why
and BO = 2/3BP . (Notice that each median is divided into segments whose lengths are
in the 2 : 1 ratio.)
(a)
(b)
Figure 1-55
Notice that if we prove first that any two medians intersect in a point whose distance
to a vertex is two-thirds of the median from that vertex, the concurrency of the medians
will follow immediately. Indeed if the medians AM and CN intersect at O as shown in
Figure 1-55 (b), then AO ∼
= 2/3AM . If AM and BP intersect at O1 , then AO1 ∼
= 2/3AM .
This implies that AO = AO1 and hence that O = O1 .
The point of concurrency of the medians of a triangle is called the centroid. It can
be shown that the centroid of a triangle is the center of gravity of the triangle when a
triangle is considered as a thin plate of uniform density. (The center of gravity can be
defined mathematically. Almost any calculus text introduces the definition.)
NOW PROVE THIS 1-7: The part of Theorem 1.31 that says that any two
medians intersect at a point that divides each median into segments whose lengths are
in the 2 : 1 ratio can be proved in way suggested in Figure 1-56. Write a proof based
on this figure. (D is the midpoint of CM and E is the midpoint of BM .)
Figure 1-56
60
GSP 1-6
Example 1.4. What kind of quadrilateral is obtained when the midpoints of consecutive
sides of the quadrilateral are connected?
Figure 1-57
Pause,
Conjecture
& Prove
Solution. In Figure 1-57 the midpoints M , N , P , and Q of the sides of the quadrilateral
ABCD are connected to form the quadrilateral M N P Q. Starting with different quadrilaterals, accurate constructions reveal that in each case M N P Q is a parallelogram. This
fact can be proved for all quadrilaterals (and actually discovered without experimentation)
as follows.
The fact that M and N are midpoints suggest the use of the Midsegment Theorem. The
side M N is a midsegment in "ABC. Thus: (1) M N ||AC and M N ∼
= 1/2AC. Similarly
in "ADC: (2) QP ||AC and QP ∼
= 1/2AC.
Now (1) and (2) imply that M N ||P Q and M N ∼
= P Q. This happens if and only if
M N P Q is a parallelogram.
Example 1.5. Complete each of the following statements assuming that the quadrilateral
M N P Q is the qudrilateral defined in Example 1.4. (Prove you answers),
Pause &
Answer
(1) M N P Q is a rhombus if and only if ABCD is a quadrilateral such that:
.
(2) M N P Q is a rectangle if and only if ABCD is a quadrilateral such that:
.
Solution. (1) Using the result of Example 1.4, M N QP in Figure 1-58 is a rhombus
if and only if M N ∼
= M Q. We need to write an equivalent condition involving
characteristics of ABCD. We have M N = 1/2AC and M Q = 1/2BD. Thus
M N = M Q if and only if 1/2AC = 1/2BD, i.e., AC ∼
= BD. Consequently, M N QP
is a rhombus if and only if the diagonals of ABCD are congruent.
(2) It is straightforward to show that if ABCD is a rhombus or even a kite, then M N P Q
is a rectangle. However, neither condition is necessary; M N QP can be a rectangle
if ABCD is not a kite (recall that a rhombus is a special case of a kite). To find
61
Figure 1-58
an “if and only if” condition we can proceed as follows. Since M N QP is always
a parallelogram it will be a rectangle if and only if one of its angles is 90◦ . This
will happen if and only if M Q⊥M N . Because M Q+BD and M N +AC, the sides
M Q and M N will be perpendicular if and only if the corresponding diagonals are
perpendicular. (Why?) Consequently statement (2) could be completed by: its
diagonals are perpendicular.
Example 1.6.
(1) State and prove the midsegment theorem for trapezoids.
(2) Prove that the segment connecting the two midpoints of the diagonals of a trapezoid
is parallel to the bases of the trapezoid and find the length of the segment if the
lengths of the bases are a and b.
Solution.
(1) Midsegment Theorem for Trapezoids:
The midsegment of a trapezoid (the segment connecting the midpoints of the sides)
is parallel to the bases and its length is 1/2 the sum of the length of the bases.
Figure 1-59
62
Given: In figure 1-59, BC+AD, M and N are respectively midpoints of AB and
CD. Prove: M N +AD and M N = 1/2(AD + BC).
Proof. Through M construct line ! (not shown in the figure) parallel to the bases. A
property of parallel projections implies that ! intersects CD at its midpoint. Because
midpoint of a segment is unique it follows that ! intersects CD at N . Consequently
←−→
M N is the line ! and is therefore parallel to the bases.
To find the length of M N given the lengths of the bases, a and b, consider the two
triangles created by the diagonal BD (the other diagonals could be used equally
well). We have M E = a/2 and EN = b/2. Hence M N = (a + b)/2.
Pause &
Consider
Why
(2) Given: In Figure 1-60, P and Q are the midpoints of the diagonal. Prove and
find: P Q is parallel to the bases of the trapezoid and express P Q in terms of a and
b.
Figure 1-60
Proof. We try to make P Q a midsegment of a triangle. One such triangle can be
obtained by connecting C with Q. The line CQ intersects AD at F . We show that
P Q is a midsegment in "ACF . Because P is the midpoint of AC (given), it remains
to show that CQ ∼
= QF . For that purpose notice that by ASA (Why?) we have:
"BCQ = "DF Q.
Thus CQ ∼
= QF and therefore P Q is a midsegment in "ACF . Consequently we
have: P Q||AF and P Q = 1/2AF . We now express AF in terms of a and b. P Q =
1/2AF = 1/2(AD − F D). We have AD = a and F D = BC = b. Hence P Q =
1/2(a − b).
63
NOW PROVE THIS 1-8: You can now prove your conjecture about the location of the treasure described in The Treasure Island Problem in Chapter 0 (see
also Problem 0-1 of Problem Set 0). You must have conjectured that the treasure is
independent of the location of the gallows Γ in Figure 1-61.
i. Show that if the conjecture is true then by choosing the gallows Γ at T2 (tree
number 2) it follows that the treasure M is on the perpendicular bisector of the
segment T1 T2 connecting the trees T1 and T2 , half the distance between the trees
above the lines connecting the trees.
ii. Use the ideas in the figure to prove the assertion in part (i). (We labeled ΓC = h,
S1 A = a and S2 B = b. You need to show that T1 A = T2 B = h, CT1 = a and
CT2 = b. Also that M N = 1/2 T1 T2 and T2 N = T1 N .)
Figure 1-61
Problem Set 1.4
In the following problems if not stated otherwise the Parallel Postulate may be used.
1. Which part of the following statement requires the use of the Parallel Postulate and
which does not. Justify your answer.
“Two lines cut by a transversal are parallel if and only if a pair of corresponding angles are congruent.”
2. (a) In Now Prove This 1-4 you have found the sum of the measures of the interior
angle in a convex quadrilateral. Is your answer true for concave quadrilaterals
64
as well?
(b) Find the sum of the measures of the interior angles of any convex n-gon in
terms of n by drawing all the diagonal from one vertex as shown below. (You
may have used this approach in Now Try This 1-4.)
∗ (c)
Prove your answer to part (b) by Mathematical Induction.
3. Justify your conjecture in Now try This 1-4 about the sum of the measures of the
angles of any convex n-gon by choosing any point P in the interior and connecting
P with each of the vertices as shown.
−−→ −−→ −−→ −−→
4. If m(∠ABC) = α and m(∠A" B " C " ) = β, BA+B " A" , BC+B " C " as in the figure below,
how are α and β related? Prove your answer.
65
5.
(i) Let m, n, a and b be lines such that m+n, m ⊥ a and n ⊥ b. Prove that a+b.
(ii) Use part (i) to prove that any two perpendicular bisectors of two sides in a
triangle must intersect.
(iii) Did the proof of (i) (and hence of (ii))depend on the parallel postulate? Justify
your answer.
6. (a) In the accompanying figure the measures of the indicated angles are as shown.
Complete the following and prove the completed statement.
a||b if and only if x, y, and z satisfy the condition
.
(b) Write and prove a statement similar to the one in part (a) that corresponds to
the figure below.
(c) Write and prove a statement similar to the one in part (b) for
(i) five angles
(ii) six angles
(d) Generalize the statement in part (c) for n angles α1 , α2 , . . . , αn .
66
GSP 1-7
7. In the figure shown a||b. The angles at P and Q have been trisected. The trisection
rays form a quadrilateral ABCD. What is the most that can be said about the
angles of ABCD? Prove your answer.
8. In "ABC the angle bisectors at B and C form "BDC. If m(∠A) = α, can m(∠D)
be found in terms of α? (If so find it. If not prove that it cannot be found.)
9. Two exterior angles in "ABC have been bisected as shown. Can m(∠D) be found
if the measures of the angles of "ABC are not known? (If so find it. If not show
that m(∠D) cannot be determined.)
10. Complete each of the following and then prove the completed statements.
(a) A quadrilateral is a rhombus if and only if its diagonals.....
67
(b) A quadrilateral is a square if and only if its diagonals.....
(c) A quadrilateral is a rectangle if and only if its diagonals .....
*(d) A quadrilateral which is not a parallelogram is an isoceles trapezoid (a trapezoid
in which the non-parallel sides are congruent) if and only if its diagonals .....
11. The following diagram suggests a proof of Theorem 1.13 that a > b ⇒ α > β. Based
on these diagrams write a proof of Theorem 1.13.
(a)
(b)
Figure 1-62
12. Prove that in a 30◦ − 60◦ − 90◦ triangle the side opposite to the 30◦ angle is half as
long as the hypotenuse. (Hint: Such a triangle is “half” an equilateral triangle.)
13. Devise as many different methods as possible for constructing a line through a given
point parallel to a given line. Justify each method.
14. If k and ! are lines with transversal m and f : k → ! is a parallel projection, explain
why f is a one-to-one and onto function.
15. Divide a segment into five congruent parts. Justify your construction.
16. Give an alternate proof of the Midsegment Theorem by extending M N through P ,
drawing CP ||AB, and proving that M P CB is a parallelogram.
68
17. Give an alternate proof of the fact that any two medians divide each other in the
ratio 2 : 1 by using the following suggestion: If P and Q are the midpoints of the
medians AO and CO respectively then M N QP is a parallelogram.
18. Let ABCD be a quadrilateral and M N QP the quadrilateral obtained by connecting
the consecutive midpoints of the sides of ABCD. Complete the following and prove
the completed statement.
M N QP is a square if and only if ABCD has the property
.
19. Give an alternative proof of the Midpoint Theorem for Trapezoids (in Example 1.6)
by considering a proof suggested by tracing the trapezoid ABCD, turning the traced
trapezoid “upside down”, and attaching the two so that their bases are collinear.
20. What, if anything, is missing (or wrong) in the following proof of Example 1.6 (2)?
Through P draw a line parallel to BC intersecting AB at S. A property of parallel
projection implies that S is the midpoint of AB. Applying the Midsegment Theorem
in "ABD and "ABC we have: P Q = QS − P S = 1/2AD − 1/2BC = 1/2(a − b).
21. (a) Let P and Q be the midpoints of the diagonals as in problem 17. Draw BP and
CQ as shown and express U V in terms of a and b where AD = a and BC = b.
Justify your answer.
(b) Under what condition will U = V ?
69
22. What kind of quadrilateral is obtained by connecting the intersection points of the
angle bisectors of a rectangle? Tell the most you can about the obtained quadrilateral
and prove your answer.
23. What kind of quadrilateral is always obtained by connecting the intersection points
of the angle bisectors of the following figures. Tell the most you can about the
quadrilaterals obtained and prove your answers.
(a) A parallelogram.
(b) An isosceles trapezoid.
24. (a) Prove that in a right triangle the median to the hypotenuse is half as long as
the hypotenuse.
(b) State and prove a converse of the the statement in (a).
70
25. In the figure below, ABCD is a square and all marked angles are congruent. What
is the most that can be said about quadrilateral EF GH? Prove your answer.
26. Prove that the converse of the Midsegment Theorem, that is: If a segment connects
two points on two sides of a triangle in such a way that it is parallel to the third side
and is half the length of that side, it is a midsegment.
27. ABCD is a parallelogram whose diagonals intersect at E. The distance of the points
A, B, C, D, and E from ! are a, b, c, d, and e respectively.
(a) Prove that b + d = a + c.
(b) Find e in terms of a, b, c, and d.
28. The segment AB (of fixed length) is moving so that the endpoint A and B are on
the sides of a right angle. What is the locus (set of points) of all the midpoints of
AB? Prove your answer.
71
29. (a) Prove that the 3 midsegments of a triangle divide it into 4 congruent triangles.
(b) Construct and acute angle triangle and its 3 altitudes.
(c) Construct an obtuse angle triangle and its 3 altitudes. Extend some of the
altitudes to show that the lines containing the altitudes are concurrent.
(d) Prove that the altitudes of a triangle or the lines containing the altitudes are
concurrent. (Hint: Through each vertex of the triangle construct a line parallel
to the opposite side. The three lines intersect in three points determining a new
triangle. Show that the perpendicular bisectors of the sides of the new triangle
contain the altitudes of the original triangle.)
∗ 30.
In the following figure three adjacent squares form a rectangle. Prove that x+y+z =
90◦ .
31. For which point P in the interior of an equilateral triangle "ABC will the sum of the
distances from the three sides be minimum. (The sum of the distance is x + y + z.)
Hint: Try various points and measure the corresponding value of x + y + z. Make a
conjecture and prove the conjecture first for points on a side of a triangle.
32. Describe how to perform each of the following constructions with “obstructions.”
(a) Construct the perpendicular bisector of a segment which is near the bottom of
a page and hence no drawing below the segment is allowed.
72
*(b) Construct a part of the line containing the angle bisector of an angle whose
vertex is not on the page.
33. Is the Parallel Postulate (Axiom 2) equivalent to the following? (Justify your answer.) Two intersecting straight lines cannot be both parallel to the same straight
line.
34. The following is a sketch of a proof showing that the Parallel Postulate implies
Euclid’s Fifth Postulate. Complete the details of the proof.
Lines !1 and !2 intersect a transversal k at points A and B respectively as
shown. Assume that α + β < 180◦ . If β1 is supplementary to α then we
can conclude that β1 > β. Hence there exists line !3 through B that forms
with k an angle whose measure is β1 as shown. Hence !3 ||!1 . Because
β1 > β, !2 (= !3 . Consequently !1 and !3 are not parallel and must meet.
This will contradict the Exterior Angle Theorem.
∗ 35.
Prove that if ABCD is a quadrilateral in which M N = 12 (a + b), where M is the
midpoint of AB, N the midpoint of CD, AD = a and BC = b, then ABCD is a
trapezoid.
73
∗ 36.
In the Treasure Island Problem Prove that the location of the treasure is independent
of the location of the gallows by considering two positions for the gallows, one at
one of the tress Γ∗ and the other Γ in an arbitrary position. (Hint: Prove that the
segment S1 S2 and S1∗ S2∗ bisect each other.)
74
Chapter 2: Circles
A circle is the set of all points (locus) in the plane, at a given distance (the radius) from
a fixed point in the plane, the center. Two circles are congruent if and only if their radii
are equal. The circle is one of the most commonly encountered figures. Wheels, balls,
cross sections of trees, the shape of the full moon, and the equator all suggest circles.
Some of the most important concepts in mathematics like π and the circular functions
(trigonometric functions) are related to the circle. Circles hold fascinating properties
which will be explored in this chapter.
2.1
Basic Properties of Arcs, Central and Inscribed Angles
A line that intersects a circle in two points is called asecant and the segment connecting
←→
the two points is a chord. In Figure 2-1(a), P Q is a secant and P Q is a chord. A chord
containing the center of the circle is a diameter, which is twice the radius.
Remark. Notice that the terms radius and diameter are used both for the segments and
their lengths. When we refer to length, we will use the prefix “the”. Thus a radius is a
segment from the center to a point on the circle. The radius is the length of any such
segment.
←→
The secant P Q divides the circle into two arcs. The larger of the two arcs is called the
!
major arc and the smaller the minor arc. In Figure 2-1(a), P XQ (read arc P XQ) is
!
the minor arc and P Y Q is the major arc. We will denote the minor arc connecting P and
!
Q by P Q.
An angle whose vertex is the center of the circle is a central angle. In Figure 2-1(b),
!
∠P OQ is a central angle whose sides intersect the circle at P and Q. The arc P Q is in the
!
!
interior of that angle. We say that ∠P OQ intercepts P Q, and P Q subtends ∠P OQ.
!
We also say that chord P Q subtends P Q.
It is convenient to associate to a minor arc the measure of its central angle.
!
If
m(∠P OQ) = 100◦ , we may write m(P Q) = 100◦ . Thus, the measure of a minor
arc is defined to be the measure of its central angle. (Another measure of an arc is its
length, which will be discussed later.) The measure of a major arc is defined to be
360◦ minus the measure of the minor arc. If P Q is a diameter as in Figure 2-1(c), then
the two arcs formed are called semicircles and the measure of each is 180◦ .
75
Figure 2-1
Adjacent arcs in a circle are arcs that have exactly one point in common. In Figure 2!
!
1(a), the arcs P X and XO are adjacent. It follows from the Angle Addition Postulate
(Appendix I) that the measure of an arc which is the union of two adjacent arcs is the
sum of the measures of these two arcs. Congruent arcs are arcs in the same circle or in
congruent circles that have the same measure.
The following propositions are intuitive. Their proofs are straight forward and hence
left out.
Proposition 2.1. Two minor arcs in the same circle or in congruent circles are congruent
if and only if their central angles are congruent.
Proposition 2.2. In a circle, or in congruent circles, two arcs are congruent if and only
if the corresponding chords are congruent.
Proposition 2.3. A diameter perpendicular to a chord bisects the chord and bisects each
of the arcs determined by the chord.
!
We illustrate Proposition 2.3 in Figure 2-2 where CD⊥AB and hence CD bisects BDA
!
and BCA (at D and C respectively).
76
Figure 2-2
In Figure 2-3, CD is the common chord of the circles with centers at A and B. The line
←→
AB is the perpendicular bisector of CD. We state this fact in the following proposition.
Proposition 2.4. If two circles intersect, the line through their centers is the perpendicular bisector of their common chord.
Figure 2-3
Proof. Because BC ∼
= BD and AC ∼
= AD, ABCD is a kite and the proposition follows
from the properties of a kite.
Tangent to a Circle
A tangent at a point P to a curve can be roughly described as a line containing P and
having the same direction as the curve. More precisely, if Q is any point on the curve,
←→
QP is a secant. If Q gets closer and closer to P (from either side of P ), the secant often
approaches a fixed line t. That fixed line (if it exists) is called the tangent to the curve at
t. A precise definition of a tangent involves the concept of a limit.
77
Figure 2-4
In the case of a circle it seems that the secant in Figure 2-5 approaches the line t which
is perpendicular to the radius P O. This observation is the basis of the following definition
which applies only to circles.
Figure 2-5
Definition of a Tangent to a Circle
A tangent to a circle O at point P on the circle is the line through P ,
perpendicular to the radius OP .
NOW PROVE THIS 2-1: A tangent to a curve may intersect the curve at
more than one point. However, in the case of a circle, the tangent intersects the circle
at exactly one point. Prove this fact by choosing any point T on the tangent and
showing that the distance from the center to T is greater than the radius.
78
An Angle Formed By a Tangent and a Secant
At any given moment an object moving along a circular path is “facing” in the direction
of the tangent to the circle. (To see why, consider an object moving from P to Q in
←→
Figure 2-5. If Q is very close to P , P Q approximates the tangent at P .) Suppose the
particle in Figure 2-6 moves along the circle from point P to point Q.
Figure 2-6
Notice that at Q the particle is “facing” in the direction of the tangent t" at that point.
!
The particle moved along the arc P Q whose measure is β, the measure of the central
angle P OQ. If we connect P with Q an angle between the tangent t and the chord P Q is
formed. We denote the measure of that angle by α. It seems that the size of α depends
on the size of β. Notice that if β = 0, then α = 0 and if β = 180◦ , then P and Q are
diametrically opposite and α = 90◦ . In fact we can prove that α = 1/2 β in all cases as
stated in the following Theorem.
Theorem 2.1. The measure of the angle between a tangent and a chord equals half the
measure of the intercepted arc.
Restatement: Referring to Figure 2-6 we need to prove the following:
(2.1)
(2.2)
α=
m(∠SP Q) =
1
2
1
β
2
(If α is acute),
!
of measure (major arc P Q).
Notice that (2.1) is the statement of the theorem for the acute angle α while (2.2) is the
theorem for an obtuse angle between a tangent and a chord.
79
Proof. To find a relationship between α and β in Figure 2-6 we use the fact that t⊥OP
and the fact that the sum of the measures of the angles in any triangle is 180◦ . Because
∆P OQ is isosceles and m(∠OP Q) = 90◦ − α we have: (90◦ − α) + (90◦ − α) + β = 180◦ .
Consequently α = 1/2β. To prove part (2.2) we express m(∠SP Q) in terms of α and
use part (2.1). Hence: m(∠SP Q) = 180◦ − α = 180◦ − 1/2 β = 1/2(360◦ − β). Because
!
1/2 (360◦ − β) is the measure of the major arc P Q, we have proved (2.2).
Figure 2-7
NOW PROVE THIS 2-2: Figure 2-7 suggests a quick way to prove Theorem 2.1
if the angle formed by the chord and tangent is acute. Supply the details of the proof
showing that α = 1/2 β.
The Inscribed Angle Theorem
Figure 2-8
80
Theorem 2.1 can be used to deduce a similar relationship between the measure of an
angle formed by two chords intersecting at a point P on a circle (see Figure 2-8) and the
intercepted arc. Such an angle is called an inscribed angle. Thus an inscribed angle is
an angle whose vertex is on the circle and whose sides are chords of the circle. Based on
this discussion we pose the following problem.
Problem 2.1. Find a relationship between the measure of an inscribed angle and the
measure of its intercepted arc.
Figure 2-9
!
Solution. In Figure 2-9, we need to find a relationship between α and m(AB). In order
←→
to use Theorem 2.1 we draw the tangent P Q. Each of the angles that the tangent makes
with P A and P B can be expressed now in terms of the corresponding measures of the
intercepted arcs. In addition, α is the difference of these angles. Referring to Figure 2-9
we have:
α2 =
!
1
m(P AB)
2
α1 =
!
1
m(P A)
2
81
α = α2 − α1 =
!
1
m(P A)
2
α=
!
!
1
[m(P AB) − m(P A)]
2
α=
!
1
m(AB).
2
Notice that the above solution does not depend on the size of α and hence is completely
general. We summarize the result of Problem 2.1 in the following useful theorem.
Theorem 2.2 (The Inscribed Angle Theorem). The measure of an inscribed angle equals
half the measure of the intercepted arc.
Notice that the proof of Theorem 2.2 is given in the solution to Problem 2.1. Figure 210 illustrates the theorem in the case where the center of the circle is in the interior of
the inscribed angle. A different proof of Theorem 2.2 (common in high-school textbook)
which does not use Theorem 2.1 will be explored in the problem set.
Figure 2-10
The Inscribed Angle Theorem is fundamental in the study of circles as we shall use it
frequently. Its full significance is perhaps more evident in the following corollaries.
Corollary 2.1. In any circle all the inscribed angles intercepting the same arc are congruent.
Proof. By Theorem 2.2 the measure of each of the inscribed angles in Figure 2-11 is half
!
the measure of BC.
Theorem 2.2 along with Corollary 2.1 is usually referred to as the Inscribed Angle
Theorem. ∠BAC in Figure 2-11 is sometimes referred to as the angle by which the chord
82
Figure 2-11
BC is seen from point A. We say that BC is “seen” from A by ∠BAC. Using this
!
terminology the Inscribed Angle Theorem says that from all points on BAC the chord BC
is seen by the same angle, whose measure is half the measure of the intercepted arc.
In Figure 2-11 the measure of ∠BAC is designated by α. We have shown that from
!
every point on BCA the chord BC is seen by angle α. However we have not proved that
!
←→
the points on BAC are the only points in the half plane determined by BC and A (the
←→
half plane “above” BC) from which BC is seen by α. In the next theorem we show that
this is the case, that is, that there are no other points in the half plane from which BC is
seen by α.
!
Theorem 2.3. If α is the measure of an inscribed angle subtending an arc BC of a circle,
then the locus1 of all the points (in the half plane determined by ↔ BC and point A on
!
the circle) from which BC is seen by α, is the arc BAC.
Figure 2-12
1
The locus is the set of all points satisfying a given condition.
83
!
Proof. In Figure 2-12(a) let m(∠A) = α. We need to show that if P is not on BAC but
in the mentioned half plane then m(∠BP C) (= α. We distinguish two cases: one where
P is in the interior of the circle and one where P is in the exterior. Figure 2-12(a) shows
←→
a point P in the interior of a circle and in the half plane determined by BC and A. Let
−−→
m(∠BP C) = β. It seems that β > α. To prove this we extend BP until BP intersects
the circle at Q as shown in Figure 2-12(b). In this way we create ∠Q whose measure is
also α (Corollary 2.1). In ∆P QC, β is an exterior angle and hence β > α. The case when
P is in the exterior of the circle is similar and is left for the reader to prove in the next
Now Try This.
NOW PROVE THIS 2-3:
1. Prove Theorem 2.3 for the case when point P is in the exterior of the circle.
2. Given a segment, construct the locus of all the points in the plane from which
the segment is seen by an angle of (a) 45◦ (b) 30◦ .
In Figure 2-13, each of the marked angles is inscribed in a semicircle. What can be
said about such inscribed angles? Because the measure of a semicircle (as an arc) is 180◦
each of the inscribed angles measures 1/2 · 180◦ or 90◦ . Hence:
Figure 2-13
Corollary 2.2. Any angle inscribed in a semicircle is a right angle.
Remark. An angle inscribed in a circle is a right angle if and only if it subtends a
diameter.(Why?)
84
Example 2.1. In Figure 2-14 AO⊥OC, O is the center of the circle, DE is the perpendicular bisector of OC. Find α and β where α and β are angles inscribed in circle O as
shown in Figure 2-14.
Figure 2-14
Solution. α, β, and α + β are inscribed angles. Hence, each equals half the measure of
!
the corresponding intercepted arc. Because AO⊥OC, we have m(AC) = 90◦ and hence:
α+β =
!
1
m(AC),
2
α+β =
1
· 90◦ ,
2
α + β = 45◦ .
Consequently, it is sufficient to find either α or β. To find α, we need only to know the
!
measures of DC or ∠DOC. The piece of information that we have not used yet is the
fact that DE is the perpendicular bisector of OC. That information implies that D is
equidistant from O and C, i.e. OD = DC. Because OD and OC are radii, ∆ODC is
!
equilateral and thus m(∠DOC) = 60◦ . Because α and ∠DOC subtend the same arc DC,
we have, α = 12 m(∠DOC) = 12 · 60◦ = 30◦ .
Hence α = 30◦ and since α + β = 45◦ , β = 15◦ .
Quadrilaterals Inscribed in a Circle
When each vertex of a polygon is on a circle like in Figure 2-15 the polygon is said to be
inscribed in a circle or in short cyclic. The circle is circumscribing the polygon.
In Chapter 1 we have seen that given a triangle there always exist a circle that circumscribes it, that is, that every triangle is cyclic. Given a quadrilateral, does there
always exist a circle that circumscribes it? Since there is a unique circle through three
85
Figure 2-15
non-collinear points, given three such points A, B, and C shown in Figure 2-16(a), we
could choose a point D not on the circle that contains A, B, and C. Because there exists a
unique circle through A, B, and C, there is no circle that circumscribes ABCD. It should
also be visually evident that a quadrilateral like the one in Figure 2-16(b) in which two
nonconsecutive vertices are relatively close to each other, is not cyclic (there is no circle
that circumscribes it).
Figure 2-16
What are the necessary and sufficient conditions for a quadrilateral to be cyclic?
Clearly if a quadrilateral like the one in Figure 2-17 is cyclic, the center of the circumscribing circle is equidistant from each vertex and hence on the perpendicular bisector of each
86
side. Conversely, if for a given quadrilateral the four perpendicular bisectors of its sides
intersect in a single point, that point is the center of the circumscribing circle. Thus: A
necessary and sufficient condition for a quadrilateral to be cyclic is for the perpendicular
bisectors of its sides to be concurrent.
However, this condition is seldom useful as there is usually no simple way to recognize
whether the perpendicular bisectors of the four sides are concurrent. (A rectangle is an
exception. Why?) Because the four angles of a cyclic quadrilateral are inscribed angles
and inscribed angles have special properties it makes sense to look for a relationship among
the angles of a cyclic quadrilateral.
Figure 2-17
In Figure 2-17 we have:
!
1
m(DB)
2
!
1
m(∠C) = m(DAB).
2
m(∠A) =
!
!
Because the union of DCB and DAB is the entire circle, we get:
m(∠A) + m(∠C) =
1
1
m(entire circle) = · 360◦ .
2
2
Our investigation implies that the sum of the measures of any pair of opposite angles in
a cyclic quadrilateral is 180◦ . Thus a necessary condition for a quadrilateral to be cyclic
is that the sum of the measures of a pair of opposite angles is 180◦ . It turns out that
the condition is also sufficient for a quadrilateral to be cyclic, as stated in the following
theorem.
Theorem 2.4. A necessary and sufficient condition for a quadrilateral to be cyclic is that
the sum of the measures of a pair of opposite angles is 180◦ .
87
Proof. We have shown above that the condition is necessary. We need to show that the
condition is sufficient. Suppose that ABCD in Figure 2-18 is a quadrilateral in which:
(2.3)
m(∠A) + m(∠C) = 180◦ .
We assume that ABCD is not cyclic and obtain a contradiction. We construct a circle
with center O that contains any three of the vertices. In Figure 2-18, O is the circle
through B, C, and D.
Figure 2-18
Since we assumed that ABCD is not cyclic, A is not on the circle. We distinguish two
cases and for each obtain a contradiction. If A is in the interior of the circle extend, AB
so that it intersects the circle at A1 . In this way we get quadrilateral A1 BCD inscribed
in a circle and hence by the already proved first part of the theorem we have:
(2.4)
m(∠A1 ) + m(∠C) = 180◦ .
Equations (2.3) and (2.4) imply that m(∠A) = m(∠A1 ), which is impossible as ∠A is an
exterior angle in ∆DAA1 , and hence greater than any non-adjacent interior angle of the
triangle. A similar contradiction is obtained for the case when A is in the exterior of the
circle and is left as an exercise.
NOW PROVE THIS 2-4: Use Theorem 2.3 to show that the condition m(∠A)+
m(∠C) = 180◦ is sufficient for a quadrilateral ABCD to be cyclic by showing that
point A in Figure 2-18 must be on the circle determined by B, C, and D.
88
NOW PROVE THIS 2-5: If available use a computer and software such as
The Geometer’s Sketchpad to perform the following constructions. (A compass and
straightedge will also suffice.)
(a) Construct a circle and a quadrilateral ABCD with perpendicular diagonals inscribed in the circle as in Figure 2-19(a). Through P , the point of intersection
of the diagonals, draw a line perpendicular to one of the sides. The line divides
the opposite sides into two segments of length a and b. Measure the segments
and find the ratio between their lengths. Repeat the experiment for few other
such quadrilaterals. Make a conjecture about your results and prove it.
(b) Draw any quadrilateral ABCD and its angle bisectors as shown in Figure 219(b). The points of intersection of the four angle bisectors form a new quadrilateral W XY Z. Construct a circle through any three of the vertices of the new
quadrilateral. What do you observe about the circle? Repeat the experiment
for two other quadrilaterals. Make a conjecture about your results and prove it.
Figure 2-19
Problem Set 2-1
In the following problems if a point is labeled by the letter O,
it is the center of the circle.
1. (a) State and prove an “if and only if” condition relating two congruent chords in a
circle to their distances to the center of the circle.
89
(b) Given two chords in a circle prove that the longer chord is closer to the center
of the circle.
(c) State and prove the converse of the statement in (b).
2. State and prove the converse of Proposition 2.3.
3. (a) Prove that a tangent to a circle intersects the circle in only one point.
(b) Prove that a line that intersects the circle in exactly one point is a tangent to
the circle.
4. Give an alternate proof of The Inscribed Angle Theorem, by first proving a special
case stated in (a) below and then using part (a) to prove the cases in (b) and (c).
In each case show that α = 1/2 β.
(a) One side of the inscribed angle contains the diameter of the circle.
(b) The center of the circle is in the interior of the angle. (Hint: construct the
diameter through the vertex of the angle and apply part (a).)
(c) The center of the circle is in the exterior of the angle.
(a)
(b)
(c)
5. Which proof of the Inscribed Angle Theorem do you like better, the one in the text
or the one outlined in Problem 5? Explain why.
6. Recall that in the text the Inscribed Angle Theorem followed from Theorem 2.1.
However as suggested in Problem 4, Theorem 2.2 can be proved without the use of
Theorem 2.1. Show that Theorem 2.1 follows from the Inscribed Angle Theorem by
viewing a tangent as a limit of secants as follows. In the figure below assume that if
P moves towards A, then AP approaches the tangent AB and in the limit ∠P AQ
approaches ∠BAQ.
90
7. In each of the following if possible find x, or x, y, and z. If not possible explain why
not. (O is the center of the circles in (a) and (c).)
(a)
(b)
(c)
8. ∆ACB is inscribed in a circle. H is the orthocenter (the point where the altitudes
−−→
intersect) of ∆ACB. Point K is the intersection of AD and the circle. Prove that
HD ∼
= DK.
91
9. (a) In the accompanied figures two secants intersect in the interior and the exterior
of the circle respectively. In each case find the angle α formed by the secants in
terms of the measures of the intercepted arcs. State and prove the corresponding
results. (Hint: Connect some of the points in the figures to create inscribed angles.)
(b) Explain why the Inscribed Angle Theorem is a special case of each of the results
in (a).
10. Prove that two parallel chords in a circle intercept congruent arcs. (If BC + AD
show AB ∼
= CD.)
11. Prove each of the following:
(a) A parallelogram is cyclic if and only if it is a rectangle.
(b) A trapezoid is cyclic if and only if it is isosceles.
92
12. ∆ABC is inscribed in a circle. Line m is tangent to the circle at C, n + m. Line n
intersects ∆ABC at D and E. Is ABED cyclic? Either prove or disprove.
13. ∆ABC is inscribed in circle O as shown. The altitude BH intersects the diameters
through A and C at D and E respectively.
(a) Prove that ∆DOE is isosceles.
(b) Is the assertion in (a) still true if ∆ABC is an obtuse triangle?
−−→
←
→
14. AB is a diameter of the semicircle with center O. Let S be on AB and SC be a
−→
tangent to the circle. If ST bisects ∠ASC find the measure of ∠ST C. (Prove your
answer.)
15. A is a point in the exterior of a circle. (Choose A close to the circle.) Let B and C
be the points of tangency of the tangents drawn from A to the circle, and let CP
93
←→
←→
be a diameter. The line P B intersects CA at point E. Prove that m(∠BAC) =
2m(∠BEA).
−−→
16. ABCD is inscribed in circle O. AB is a diameter and E is the intersection of AD
−−→
and BC.
(a) Prove that ∠ODC ∼
= ∠AEB.
(b) OD is the tangent at D to the circle that circumscribes ∆DEC.
(c) Which of the above statements, if any, are true if AB is not a diameter? Justify
your answer.
Section 2.2: Circles Inscribed In Polygons
When each side of a polygon is tangent to a circle, the circle is said to be inscribed in the
polygon and the polygon circumscribes the circle. Such a polygon is called circumscribable. Figures 2-20(a) and (b) show a circumscribable triangle and a circumscribable
quadrilateral.
(a)
(b)
Figure 2-20
94
We have seen in Chapter 1 that given any triangle, it is always possible to inscribe a
circle in it. Here is a quick review why. It seems that such a circle always exists. If that
is the case, what is the procedure for constructing it? Notice that each of the distances
OP , OQ, and OS from the center O to the corresponding sides must be equal to r, the
radius of the circle. In particular, O is equidistant from the sides of ∠A and hence must
be on the angle bisector of ∠A. Similarly, O is on the angle bisector of ∠B and of ∠C.
This suggests the following theorem already stated and proved in Chapter 1. (We state it
again for convenience.)
Theorem 2.5. The angle bisectors of any triangle intersect at a single point O which is
the center of the inscribed circle. The distance from O to any of the sides of the triangle
is the radius of the circle.
Notice in Figure 2-20(a) that P , Q, and S are the points of tangency (why?). It seems
that AP = AQ. This follows from the fact that ∆AQO ∼
= ∆AP O. Similarly, BP = BS
and CS = CP . This suggests the following useful theorem.
Theorem 2.6. The two tangent segments to a circle from a point in the exterior of the
circle are congruent.
Proof. Figure 2-21 shows the two tangent segments AP and AQ from A. We need to
show that AP ∼
= AQ. Notice that, ∆AQO ∼
= ∆AP O by Hypotenuse-Leg since the angles
at P and Q are right angles (definition of a tangent), AO ∼
= AO and P O ∼
= QO. (O is
on the angle bisector of ∠A and hence equidistant from the sides of the angle.) Hence:
AQ ∼
= AP .
Figure 2-21
95
Circles Inscribed in Quadrilaterals
It seems clear that a circle cannot be inscribed in every quadrilateral. For example,
consider the quadrilaterals in Figure 2-22(a) and (b). What makes the quadrilateral in
Figure 2-22(c) circumscribable and the other not? In Figure 2-22(c) the distance from the
center O to each of the sides of the quadrilateral is the radius of the circle.
Figure 2-22
Thus O is equidistant from the sides of any of the angles. Consequently O must be on
each of the angle bisectors, that is the angle bisectors intersect at O. Conversely suppose
that the angle bisectors of the four angles of a quadrilateral are concurrent and intersect
at point O. Can we conclude then that the quadrilateral is circumscribable? Because a
point is on the angle bisector of an angle if and only if it is equidistant from the sides of the
angle, it follows that O is equidistant from all the sides of the quadrilateral. Consequently
a circle with center O and radius equal to the distance from O to any of the sides of the
quadrilateral can be inscribed in the quadrilateral. Thus we have proved the following
theorem.
Theorem 2.7. A circle can be inscribed in a quadrilateral if and only if the angle bisectors
of the four angles of the quadrilateral are concurrent.
Notice that a circle can be inscribed in every square since the diagonals of a square
are also its angle bisectors. Can you think about other familiar quadrilaterals in which a
circle can be inscribed? (Try to find one before reading on.) Such a quadrilateral is given
in the following example.
Example 2.2. Prove that a circle can be inscribed in every kite.
96
Making a Plan.
By Theorem 2.7 it is sufficient to prove that the angle bisectors of a kite are concurrent.
Consider that kite ABCD in Figure 2-23 in which AB = BC and AD = DC.
Figure 2-23
From the properties of a kite we know that the diagonal BD bisects ∠B and ∠D. The
angle bisector of ∠A intersects BD in some point E forming two equal angles α1 and α2 .
We will show that the angle bisector of ∠C also intersects BD at E. One way to achieve
−−→
this is to connect E with C and prove that CE, (which makes angles γ1 and γ2 with the
sides DC and BC respectively), is the angle bisector of ∠C, that is to say that γ1 = γ2 .
To prove that γ1 = γ2 , we show that α1 = γ1 and α2 = γ2 . For that purpose we show
that each pair of these angles is congruent by showing that the angles in each pair are
corresponding parts in congruent triangles. Notice that ∆ABE ∼
= ∆CBE by SAS and
similarly ∆ADE ∼
= ∆CDE.
Proof. In the kite ABCD in Figure 2-23, AB = BC and AD = DC. Construct the
diagonal BD. From the properties of a kite BD bisects the angles at B and D. Next
construct the angle bisector of ∠A. Hence α1 = α2 . Let E be the point at which this angle
bisector intersects BD. Connect E with C. Notice that ∆ABE ∼
= ∆CBE by SAS since
AB = BC (definition of a kite), BE = BE and ∠ABE ∼
= ∠CBE (property of a kite).
Similarly ∆ADE ∼
= ∆CDE. Consequently ∠BCE ∼
= ∠BAE and ∠DCE ∼
= ∠DAE.
−→
Thus α1 = γ1 and α2 = γ2 . Since α1 = α2 (AE is the angle bisector of ∠A), γ1 = γ2 .
−−→
Consequently CE is the angle bisector of ∠C and therefore all the angle bisectors of a kite
are concurrent.
97
A Relationship Among the Sides of a Circumscribable Quadrilateral
In Example 2.2 we have seen that a circle can be inscribed in any kite. On the other hand
the quadrilateral in Figure 2-22(b) does not seem circumscribable. However, we can always
find a unique circle tangent to any three sides of the quadrilateral (why?), but the fourth
side is not necessarily tangent to the circle. If a circle can be inscribed in a quadrilateral
as in Figure 2-24, then all the sides are tangent to the circle and hence by Theorem 2.6 all
the tangent segments from the same vertex are congruent. This is indicated in Figure 2-24
by designating the lengths of the congruent tangent segments by the same letters.
Figure 2-24
Consequently we have:
AB = x + y
BC = z + y
DC = w + z
DA = w + x.
These equations suggest a relationship among the lengths of the four sides of the quadrilateral. Notice that:
AB + DC = (x + y) + (w + z) = x + y + z + w,
BC + DA = (z + y) + (w + x) = x + y + z + w.
Consequently:
(2.5)
AB + BC = BC + DA.
Notice that condition (2.5) was proved under the assumption that the quadrilateral is
circumscribable. Thus we have proved the following theorem.
98
Theorem 2.8. If a quadrilateral is circumscribable, then the sum of the measures of two
opposite sides equals the sum of the measures of the remaining opposite sides.
Remark. Theorem 2.8 can be stated in each of the following equivalent forms (why?):
(i) A necessary condition for a quadrilateral to be circumscribable is that the sum of the
measures of two opposite sides equals the sum of the measures of the remaining opposite
sides.
(ii) If the sum of the measures of two opposite sides in a quadrilateral does not equal
the sum of the measures of the remaining opposites sides then the quadrilateral is not
circumscribable.
Example 2.3. (i) Prove that if a parallelogram is not a rhombus it is not circumscribable.
(ii) On the basis of the theorems proved so far, can you conclude that the quadrilateral
shown in Figure 2-25(b) is circumscribable? (The sides’ measures are as shown.)
Figure 2-25
Solution. (i) If the parallelogram in Figure 2-25(a) is circumscribable then 2p = 2q or
p = q. The contrapositive of this statement says that if p (= q then the parallelogram is
not circumscribable. Thus if the parallelogram is not a rhombus it is not circumscribable.
(ii) For the quadrilateral in Figure 2-25(b) we have that the sum of the measures of opposite sides for each pair of the sides is the same since 2 + 6 = 3 + 5. However Theorem 2.8
does not tell us if condition (2.5) is satisfied, then the quadrilateral is circumscribable.
Consequently we cannot conclude yet that the quadrilateral in Figure 2-25(b) is circumscribable. For that purpose we would need to establish the converse of Theorem 2.8. This
is our next goal.
Theorem 2.9 (Converse of Theorem 2.8). If the sum of the measures of two opposite sides
of a quadrilateral equals the sum of the measures of the two remaining opposite sides, then
the quadrilateral is circumscribable.
99
Proof. Suppose that the condition of the theorem is satisfied. Referring to Figure 2-26
we have:
(2.6)
AB + DC = AD + BC.
We will show that it is possible to inscribe a circle in the quadrilateral by showing that
the angle bisectors of the four angles are concurrent.2 We first consider a special case if
two adjacent sides of the quadrilateral are congruent. Then (2.6) implies that ABCD is a
kite. (Why?) In this case, the proof is straightforward and was established in Example 2.2.
Next suppose that two adjacent sides are not congruent. Without loss of generality let
AB > AD. Then (2.6) implies:
(2.7)
AB − AD = BC − DC
and hence BC > DC.
We mark E and F on AB and BC respectively so that AE ∼
= AD and CF ∼
= DC.
Figure 2-26
Consequently, ∆DCF and ∆DAE are isosceles. Because AE = AD and DC = CF ,
substituting AE and CF for AD and DC, respectively, in equation (2.7) we get:
(2.8)
2
AB − AE = BC − CF .
The proof that follows is detailed and complete but not all the steps taken are motivated.
100
Equation (2.8) implies:
(2.9)
BE = BF .
The last equation implies that ∆BEF is isosceles. Our objective is still to show that
the angle bisectors of ABCD are concurrent. Notice that because ∆ADE is isosceles, the
angle bisector of ∠A is also the perpendicular bisector of the opposite side ED. Similarly,
the angle bisector of ∠B is the perpendicular bisector of EF and the angle bisector of
∠C is the perpendicular bisector of DF . Thus the angle bisectors of ∠A, ∠B, and ∠C
are the perpendicular bisectors of the sides of ∆DEF which we know are concurrent. It
remains to be shown that O is on the angle bisector of ∠D. This will be the case if and
only if in Figure 2-26, OS = OR. Because O is on the angle bisectors of ∠A, ∠B, and
∠C, OS = OP , OP = OQ, and OQ = OR. Consequently OS = OR.
NOW PROVE THIS 2-6: Why in the proof of Theorem 2.9 was it important
to prove first the special case when the quadrilateral is a kite?
Notice that the last two theorems combined can be stated in one theorem:
Theorem 2.10. A quadrilateral is circumscribable if and only if the sum of the measures
of two of its opposite sides equals the sum of the measures of the other two opposite sides.
←−→
←→
Example 2.4. In Figure 2-27, AB is the diameter of circle O, BM , and EF are tangents
←→
←−→
←→
(E is an arbitrary point on the circle). The tangent EF intersects BM at C and AE
←−→
intersects BM at D.
(i) Prove that BC = CD
(ii) Prove that ABCE is not circumscribable.
Figure 2-27
101
Making a Plan.
What can we conclude about C? Because CB and CE are tangent segments from point
C, by Theorem 2.6 we have:
(2.10)
BC = CE.
Notice that if BC = CD were true, then with (2.10) it would imply that CE = CD and
conversely if we could show that CE = CD then with (2.10) we would have CD = BC. It
seems easier to prove that CE = CD as these are sides in ∆ECD, and we are more familiar
with investigating triangles. In fact, it would suffice to show that the angles opposite these
sides are congruent, i.e., referring to Figure 2-27 that ε = δ. This is equivalent to what
we need to prove and seems to be easier as we have substantial information about angles
related to the circle. Thus our new subgoal is to show that ε = δ.
Notice that ε1 = ε (vertical angles). Thus it suffices to show that ε1 = δ. Since δ is
not an angle between a chord and a tangent nor an inscribed angle, we try to express δ in
terms of such angles. In ∆ABD, ∠B is a right angle (why?), hence:
δ = 90◦ − α
(2.11)
We will show that ε1 = 90◦ − α. For that purpose we connect B with E to create a
right angle BEA (since it subtends the diameter AE). Now we have ε1 = β (both angles
!
intercept the same arc AE). From the right triangle ABE: β = 90◦ −α. Thus ε1 = 90◦ −α.
This along with (2.11) implies that ε1 = δ and hence that ε = δ. Consequently, the subgoal
and therefore the required results are proved.
Proof. Because BM in Figure 2-27 is tangent to the circle at B, ∠B is a right angle. From
∆ABD, δ = 90◦ − α. Also ε1 = ε (vertical angles). Connect B with E. We have ε1 = β
!
since both angles intercept the same arc AE. Because ∠AEB subtends the diameter AB
it is a right angle. Hence from ∆ABE, β = 90◦ − α. Thus:
ε1 = 90◦ − α.
Consequently ε1 = δ and therefore ε = δ. This implies CE = CD and because
CE = BC (tangent segments from C), we have CD = BC.
In what follows we give a somewhat different proof of the result in Example 2.4. (The
Making a Plan and the proof are combined.)
Proof. II. We need to show that C is the midpoint of BD. This reminds us of the midsegment theorem. To apply that theorem we need to create a triangle and a midsegment.
For that purpose connect O with E and with C. Because BC ∼
= CE (tangent segments),
102
and OB ∼
= OE (radii), it follows that OECB is a kite. We know that the diagonals of a
kite are perpendicular to each other, that OC bisects the angles at O and C, and that M
is the midpoint of BE. We focus now on ∆ABD and try to show that OC + AD. This last
relation will follow if a pair of corresponding angles created by OC, AD, and a transversal
are congruent. Because ∠BM O is a right angle, it suffices to show that m(∠BEA) = 90◦ .
Indeed ∠BEA is a right angle as it subtends the diameter AB. Consequently OC + AD.
Because O is the midpoint of AB and OC + AD, it follows that C is the midpoint of BD
(property of parallel projections).
Which proof do you like better? Why?
Example 2.5. If ∆ABC is an equilateral triangle inscribed in a circle and P is any point
!
on AC, prove that AP + P C = P B. (See Figure 2-28.)
Figure 2-28
Making a Plan.
Referring to Figure 2-28(a) we need to show that a + b = c. Notice that c is a side in
∆ABP . Because of our experience in working with triangles, we make a + b a side in a
103
triangle. One way to achieve this is to extend P C so that DP = b as shown in Figure 228(b). Then a + b is a side in ∆ACD. Since we want to prove that DC ∼
= BP it makes
sense to look for a relationship between ∆ABP and ∆ACD. This will be the focus of our
proof.
Proof. Because ∆ABC is equilateral, each of its angles is 60◦ . Extend P C so that
DP ∼
= AP as shown in Figure 2-28(b). We show that ∆ABP ∼
= ∆ACD. Notice that:
(2.12)
∠ABP ∼
= ∠ACD
!
(inscribed angles subtending AP ) and
AB ∼
= AC
(2.13)
(hypothesis).
To prove that ∆ABP ∼
= ∆ACD it will suffice by AAS (a version of ASA) to show that
∼
∠AP B = ∠ADC.
!
Notice that ∠AP B ∼
= ∠ACB as these are inscribed angles in the same arc AE. Hence:
m(∠AP B) = 60◦ .
(2.14)
∼ ∠ADC we need only to show that m(∠ADC) = 60◦ .
Consequently to show that ∠AP B =
By our construction ∆AP D is isosceles hence m(∠ADC) = 60◦ if and only if ∆AP D is
equilateral. But ∆AP D is equilateral if and only if m(∠AP D) = 60◦ . This task is more
accessible since ∠AP D is supplementary to ∠AP C and that angle is an inscribed angle.
Also ∠ABC is supplementary to ∠AP C (opposite angles in the inscribed quadrilateral
ABCP ). Hence m(∠AP D) = 60◦ (angles supplementary to congruent angles are congruent) and consequently ∆AP D is equilateral and therefore:
m(∠ADC) = 60◦ .
(2.15)
From (2.14) and (2.15) we have:
∠AP B ∼
= ∠ADC.
(2.16)
Now from (2.12), (2.13), and (2.16) by AAS we have:
∆ABP ∼
= ∆ACD.
(2.17)
Consequently CD ∼
= BP and because CD = a + b (construction of CD) and BP = c, we
have: a + b = c.
Remark. Figure 2-28 is a vivid example of the fact that congruence of two corresponding
sides and an angle does not assure congruence of triangles. Notice in ∆ABP and ∆ACP ,
AP is a common side, AC = AB, and ∠ABP ∼
= ∠ACP , however the triangles are not
congruent since m(∠AP B) = 60◦ and none of the angles in ∆ACP is 60◦ . (Why?) Because
!
this is true for any P on AC we have exhibited infinitely many such pairs of triangles.
104
Problem Set 2-2
1. ∆ABC is inscribed in a circle. Point D is the center of the inscribed circle. Prove
that ∠DAE ∼
= ∠ADE.
2. Prove that an isosceles trapezoid with height h and bases a and b is circumscribable
if and only if h2 = ab.
3. If circle O is inscribed in a trapezoid prove each of the following:
(a) The points of tangency of the parallel bases and the center O are collinear. (In
the figure P , O, and Q are collinear.) Hence show that the diameter of the circle
equals the height of the trapezoid.
(b) In the accompanied figure prove that m(∠COB) = 90◦ .
4. Two congruent circles intersect at P and Q as shown. Through Q a line is drawn.
The line intersects the circles at A and B. The line AP intersects the circles at A
and C.
(a) Prove that P B = P A.
(b) If AQ is a diameter prove that CP = AP and that m(∠CBA) = 90◦ .
105
5. The two circles shown share a common center (concentric circles). If AB and CD
are two chords in the larger circle tangent to the smaller circle, prove that AB = CD.
!
6. In the accompanying figure P A and P B are tangents to circle O. If Q is on AB and
CD is tangent at Q prove that:
!
(a) m(∠COD) is the same regardless of the position of Q on AB.
!
(b) The perimeter of ∆P CD is the same regardless of the position of Q on AB.
106
←→
←→
7. Suppose P A and P B are tangents to a circle at A and B respectively and m(∠BP A)
= α. Answer the following:
(a) Express the marked angles at C and D in terms of α.
(b) Prove that α = m(∠C) − m(∠D).
8. ABCD is a quadrilateral with right angles at A and at D. Circle O is tangent to
the sides AB and AD at B and E respectively. The diagonal AC contains O. The
side CD intersects the circle at P and P B intersects AC at Q.
(a) Find the angles of ∆CQP .
(b) Prove that AQ = r, where r is the radius of the circle.
9. A circle is tangent at P to the side BC of the square ABCD. The vertices A and
D are on the circle as shown. The side DC intersects the circle at Q. Prove that:
(a) ∠QP A is a right angle.
107
(b) AP bisects ∠QAB.
D
Q
C
P
A
B
10. Determine whether it is possible to find a circumscribable isosceles “true” trapezoid
(in which the parallel sides are not congruent) which have the indicated property.
Justify your answers.
(a) A diagonal bisects an angle of the trapezoid.
(b) The trapezoid is cyclic.
(c) The diagonals are perpendicular to each other.
11. (a) Circle O is inscribed in a rhombus. ABCD is a quadrilateral whose vertices are
the four points of tangency. What kind of quadrilateral does ABCD seem to be?
Prove your answer.
(b) State and prove the converse of the theorem suggested in (a).
12. Circle O is inscribed in trapezoid ABCD, M N is the midsegment of the trapezoid.
Prove the following:
108
(a) M N contains O.
(b) ON = N B and OM = M A.
13. In a trapezoid ABCD, M N is the midsegment and P is a point on M N such that
P N = N B and P M = M A. Prove that:
(a) It is possible to inscribe a circle in ABCD.
(b) P is the center of the inscribed circle.
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14. Prove that the centers O1 and O2 of two disjoint circles (circles which have no common points), and the point P where the two tangents shown intersect are collinear.
15. Two circles which have only one point in common are called tangent circles. Prove
that:
(a) The common point of two tangent circles and the centers of the circles are
collinear. (Hint: Assume the contrary and use the triangle inequality.)
(b) The tangent to one of the circles at the point of contact is also tangent to the
other circle.
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16. Circles O1 and O2 are tangent at B to each other. A is any point on the common tan−→
−→
gent through B and AP and AQ are tangents to the circles O1 and O2 respectively.
Prove that AP = AQ.
17. The circles O1 and O2 are tangent at P . The line AB is tangent to the circles at A
−−→
and B respectively. If C is on AB and CP is the tangent through P prove that:
(a) AC = CB.
(b) m(∠AP B) = 90◦ .
18. The circles O1 and O2 are tangent at P . A line through P intersects the first circle
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at A and the second at B. Prove that the tangent at A to the first circle is parallel
to the tangent at B to the second circle.
19. Two congruent circles O1 and O2 intersect at A and B. A third circle with center
at B intersects O1 in C and O2 in D. (These points are in the same half plane
determined by AB.) Prove that A, C, and D are collinear.
20. The vertices of ∆EF D are on the sides of ∆ABC as shown. Construct the three
circles that circumscribe ∆ADE, ∆DF C, and ∆EBF . Repeat for differently positioned triangles EF D.
(a) Based on what you observed make a conjecture concerning the three circles.
(b) Prove your conjecture in (a).
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21. (a) Prove that whenever three circles (as shown) are tangent to each other the three
tangents at the points of contact are concurrent.
(b) Use the result in part (a) to construct three noncongruent circles tangent to each
other. Describe your construction.
" 22. We have seen that the property that all inscribed angles subtending the same arc
in a circle are congruent happens only in a circle. (See Corollary 2.1 and Problem
10.) Since both the property and its converse are true, this property of an inscribed
angle is a characteristic property of the circle. Consider the property: A chord
joining any two points on a circle meets the circle at two congruent angles. (The
angle between a line and a circle is the smaller angle between the line and the
tangent to the circle at the point of intersection. The two congruent angles are
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marked in the figure.) Prove that the property is a characteristic property of the
circle.
2.2
More on Constructions
In this section we will investigate several constructions based on the Inscribed Angle Theorem (Theorem 2.2), its corollary and
From all the points of the arc BAC (Figure 2-29) the chord BC is seen by
the same angle whose measure is half of the measure of the intercepted arc.
Moreover the arc BAC consists of all the points (in a half plane determined
←→
by BC) from which BC is seen by that angle.
Figure 2-29
These facts will be used in many of the construction problems in this section. Each
problem will include three parts:
(i) Investigation (sometimes also referred to as analysis) in which we imagine the
given problem as solved and search for properties of the figure that will enable us to
accomplish the construction.
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(ii) Construction where we describe the steps in the construction and actually perform
the construction using only compass and straight edge (or software such as GSP).
(iii) Proof where we prove that our construction does what was asked for.
Problem 1 Construct the locus (set) of all points from which a given segment a is
seen at a given angle α. The segment and the angle are given in Figure 2-30.
Figure 2-30
Investigation
We are imagining that the problem has been solved and therefore BC in Figure 2-31 is
seen by α from any point on the arc BAC. From the discussion at the beginning of this
section we know that arc BAC is the only set of points in the half plane determined by
←→
←→
BC and A for which BC is seen by α. (The reflection of arc BAC in BC is the set of
points in the other half plane from which BC is seen by α.)
Figure 2-31
In order to construct the required locus that is the arc BAC we need only find the
center O of the circle to which the arc belongs. Thus we connect in Figure 2-31 O with
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B and C. Notice that O is equidistant from B ad C and hence on the perpendicular
bisector n of BC which is also the angle bisector of ∠BOC. If we could construct line
k in Figure 2-31 we would be able to find O as the intersection point of lines n and k.
Notice that in "M BO, ∠M BO measure 90 − α which suggests drawing line BD perpendicular to BM . Then we would have m(∠DBO) = α, (this also follows from the fact that
←→ ←−→
BD = OM ). Since α is given we know now how to construct line k which contains a side
of α.
Construction (Figure 2-32)
Figure 2-32
1. On any line construct BC congruent to the given segment a.
2. Construct the perpendicular bisector n of BC.
3. Through B construct line p perpendicular to BC, (or equivalently parallel to n).
4. Let D be on line p, and E on line k (both in the same half plane). Construct ∠DBE
congruent to the given angle α.
5. The intersection BE and n is O.
6. Construct the circle with center O and radius BO. The arc BC in the half plane
←→
←→
determined by BC and O is the solution. That arc along with its reflection in BC
is the solution in the entire plane.
Proof. To show that the arc BC mentioned above is the solution it will suffice
to prove that m(∠BOC) = 2α. Notice that p+n implies that m(∠BOM ) = α.
−−→
Also m(∠COM ) = α, since OM is the angle bisector of ∠BOC. Consequently,
m(∠BOC) = 2α.