Download EUCLIDEAN GEOMETRY Contents 1. Euclid`s geometry as a theory

EUCLIDEAN GEOMETRY
ANCA MUSTATA
Contents
1. Euclid’s geometry as a theory, 2
2. Basic objects and terms, 3
2.1. Angles, 4
2.2. The circle, 4
2.3. The polygon, 5
2.4. The triangle, 5
2.5. Parallel lines, 5
3. Axioms of Euclidean geometry, 5
3.1. Convexity, 7
4. Direct consequences of the axioms, 8
4.1. Angle measures, 8
5. Tessellations, 11
6. Congruent triangles, 13
7. Triangles in parallelograms, 16
7.1. Special parallelograms, 18
8. The Pythagorean theorem, 19
9. Special lines in a triangle, 21
9.1. Angle bisector, 21
9.2. Constructing angle bisectors, 21
9.3. The perpendicular bisector, 23
9.4. Altitudes, 25
9.5. Median, 29
10. Circles, 33
10.1. Secants and tangents, 33
10.2. Arcs and angles, 35
11. Similar triangles, 38
12. The nine-point circle, 41
13. Menelaus and Ceva theorems, 42
References, 43
Index, 44
Warning: please read this text with a pencil at hand, as you will
need to draw your own pictures to illustrate some statements.
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ANCA MUSTATA
1. Euclid’s geometry as a theory
God is always doing geometry
— Plato, according to Plutarch [6]
These words suggest the reverence with which this branch of mathematics was regarded by thinkers in the ancient world. They saw geometry as
managing to extract proportions, order and symmetry from the seemingly
chaotic nature, thus making its beauty accessible to the reasoning mind. For
us as well, geometry is a bridge from visual representations of the world to
abstract logical thinking. This makes it a wonderful education tool. Indeed,
since our perception of the world is embedded in sensorial experiences, what
better way to develop a solid basis for our abstract thinking than to combine
our visual intuition with logical deductions?
It is for these reasons that an ancient geometry text has been referred to
as the most famous and influential textbook ever written. The Elements is
a collection of thirteen mathematical books attributed to Euclid, who taught
at Alexandria in Egypt and lived from about 325 BC to 265 BC. This is the
earliest known historical example of a mathematical theory based on the
axiomatic and logical deduction method.
A mathematical theory consists of
• a set of basic objects described by definitions,
• a set of basic assumptions about these objects: the axioms, and
• a set of statements derived from the axioms by logical reasoning.
– the most important of these are theorems,
– while less important statements are propositions,
– and corollaries are direct consequences of some previous statement,
– and lemmas are helpful in proving further propositions or theorems.
Each theorem, proposition or lemma consists of a hypothesis (set of assumptions), which is what we know, and a conclusion: what we have to
prove. These should be followed by a proof , meaning a chain of statements
related by logical implications, which starts from the hypothesis, combines
it with the axioms and/or statements already proven, and arrives to the
conclusion.
EUCLIDEAN GEOMETRY
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2. Basic objects and terms
All human knowledge begins with intuitions, thence passes to
concepts and ends with ideas.
— Immanuel Kant
Kritik der reinen Vernunft, Elementarlehle
Euclid’s geometry assumes an intuitive grasp of basic objects like points,
straight lines, segments, and the plane. These could be considered as primitive concepts, in the sense that they cannot be described in terms of simpler
concepts. However, Euclid attempts some definitions by means of other intuitive notions like position, breath and length, in-between. Some of these
definitions are included below in italics for your enjoyment; you do not need
to remember them, but you do need to know the definitions of more complex
objects like the circle or a polygon, or relations between them like concurrence, perpendicularity, etc. Here we loosely follow [3] in giving a rough
guide to our primitive concepts, and using them to define various related
concepts.
“A point is that of which there is no part.” A point is usually denoted by
an upper case letter.
“A straight line is length without breadth, which lies evenly with points
on itself.” A straight line is usually denoted by a lower case letter. We will
think of a line as a set of points. Write A ∈ d if A is a point on the line d.
Alternatively, we may denote a line by any two points on it: d = AB.
d
A
B
b
b
Note: C ∈ AB means that C is a point on the line AB, namely C can be in
any one of these 3 positions:
C
A
b
B
b
b
A C
b
B
b
A
b
B
b
C
b
b
A point A on a line d divides the line into two half-lines, or rays. Two
points A and B on the line d determine the segment [AB], made of all the
points between A and B. This segment may also be denoted by AB.
A
B
b
b
[AB]
If three or more lines intersect at a point, they are concurrent at that
point. If three or more points are on the same line, they are collinear.
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ANCA MUSTATA
“A surface is that which has length and breadth. When a surface is such
that the line joining any two arbitrary points in it lies wholly in the surface,
it is a plane.” A line in a plane divides the plane in two half-planes.
2.1. Angles. In a plane, consider two half-planes bounded by two lines
concurrent at the point O. The intersection of the two half-planes is an
angle. The two lines are the legs, and the point the vertex of the angle. A
\ of which
particular angle in a figure is denoted by three letters, as BAC,
the middle one, A, is at the vertex, and the other two along the legs. The
angle is then read BAC.
b
A
B
α
b
b
C
The angle formed by joining two or more angles together is their sum.
Thus the sum of the two angles ABC, P QR is the angle formed by applying
the side QP to the side BC, so that the vertex Q falls on the vertex B, and
the side QR on the opposite side of BC from BA.
When the sum of two angles BAC, CAD is such that the legs BA, AD
form one straight line, they are supplements of each other.
When one line stands on another, and makes the adjacent angles at both
sides of itself equal, each of the angles is a right angle, and the line which
stands on the other is a perpendicular to it.
b
b
b
Hence a right angle is equal to its supplement. An acute angle is one which
b
is less than a right angle.
b
b
An obtuse angle is one which is greater
b
than a right angle.
The supplement of an acute angle is obtuse,
and conversely, the supplement of an obtuse angle is acute.
When the sum of two angles is a right angle,each is the complement of
the other.
b
b
2.2. The circle. A circle is a plane figure formed by a curve, the circumference, and is such that all segments drawn from a certain point within the
figure to the circumference are equal to one another; this point is the centre.
EUCLIDEAN GEOMETRY
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A radius of a circle is any right line drawn from the centre to the circumference. A diameter of a circle is a right line drawn through the centre and
terminated both ways by the circumference. Four or more points found on
the same circle are concyclic.
2.3. The polygon. A figure bounded by three or more segments is a polygon. The segments are the sides of the polygon.
b
B
C
b
b
A
b
b
D
E
A polygon of three sides is a triangle. A polygon of four sides is a quadrilateral. A polygon which has five sides is a pentagon; one which has six sides
is a hexagon, and so on.
2.4. The triangle. A triangle whose three sides are unequal is scalene; a
triangle having two sides equal is isosceles; and, having all its sides equal,
is equilateral. A right-angled triangle is one that has one of its angles a
right angle. The side which subtends the right angle is the hypotenuse. An
obtuse-angled triangle is one that has one of its angles obtuse. An acuteangled triangle is one that has its three angles acute. An exterior angle of
a triangle is one that is formed by any side and the continuation of another
side. Hence a triangle has six exterior angles; and also each exterior angle
is the supplement of the adjacent interior angle.
2.5. Parallel lines. are straight-lines which, being in the same plane, and
being continued to infinity in each direction, meet with one another in neither (of these directions).
3. Axioms of Euclidean geometry
Postulates:
1) A unique straight line segment can be drawn joining any two distinct
points.
2) Any straight line segment is contained in a unique straight line.
3) Given any straight line segment, a unique circle can be drawn having
the segment as radius and one endpoint as center.
4) All right angles are congruent.
5) Given a point not on a given line, there exists a unique line through
that point parallel to the given line.
Common notions:
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1) Things which are equal to the same thing are also equal to one
another.
2) If equals be added to equals, the wholes are equal. In other words, if
a1 = a2 and b1 = b2 then a1 + b1 = a2 + b2 . This is true for numbers
as well as for segments and angles.
3) If equals be subtracted from equals, the remainders are equal. In
other words, if a1 = a2 and b1 = b2 then a1 − b1 = a2 − b2 .
4) Things which coincide with one another are equal to one another. In
other words, if we can move a figure (angle or segment) to fit exactly
on top of the other, then it means they are equal (in terms of size).
5) The whole is greater than the part. In other words, we can show
an object to be smaller than another object by moving the smaller
object until it fits inside the larger one, thus becoming a part of it.
The common notions can be extended to say that the segments and angles
can be measured by real numbers, and as such, they satisfy all the properties
expected of real numbers.
In truth, Euclid’s axioms are not sufficient for formally deducing the theorems of Euclidean geometry, or even for defining notions like “equal things,”
or for comparing angles and segments. Apart from the axioms, Euclid also
relied on other “common sense” intuitive notions like rigid movement, and
basic notions of topology of the plane like boundaries, interior and exterior
of a figure, and so on.
The notion of rigid movement is necessary when comparing geometric
objects. A rigid movement of a geometric figure in plane can be understood
as cutting the figure out of a sheet of paper representing the plane and
placing it in a different place in the plane. In practice, we don’t cut figures
out in order to move them – we clone them! (copy them exactly) by means
of markings on a ruler (for segments) or protractor (for angles). As a basic
tenet of Euclidean geometry, you can thus move any geometric figure found
somewhere in the plane to any other position in plane. Interestingly, Euclid
put effort into proving this tenet for movements of segments, while taking
it for granted in the case of angles. Rigid motion by means of a ruler
and protractor is so ingrained in our way of doing geometry that we don’t
even notice how the notion of measure (centimeters, meters, inches etc. for
segments, and degrees for angles) is in fact an indirect process resulting from
being able to compare and add objects by moving them in the correct places.
Two geometric figures X and Y are congruent if one can move the first
figure and superpose it exactly on top of the second figure, such that the
points of the two figures now coincide. In this case we write X ≡ Y . A
segment AB is smaller than another one CD if one can move the segment
AB until A coincides with D and B is in between C and D. Similarly, an
angle AOB is smaller than CO′ D if one can move AOB such that O falls
over O′ , the line OA over O′ C ′ , and B is in the interior of the angle CO′ D.
EUCLIDEAN GEOMETRY
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3.1. Convexity. A line cuts any plane into two sides, the half-planes. A
segment intersects the line if and only if the end-points of the segment are
on different sides of the line.
b
b
A bounded figure is convex if, for any two points A and B in the interior of
the figure, the segment AB is also in the interior of the figure.
The polygon CDEF G is convex, but HIJKLM is not convex. If a line
contains a point found in the interior of a convex figure, then it intersects
that figure in exactly two points.
b
b
Even after gathering all these extra basic assumptions in a set of axioms,
there would be some work to be done. One would have to eliminate the
superfluous assumptions, i.e. those which can be considered as theorems
or propositions based on the other axioms. For example, we do not need
to assume rigid motion for all figures—only for angles and segments. On
the other hand, Euclid proved that a segment can be moved to any other
position if we assume that two circles, each passing through the interior of
the other, intersect. Another problem may appear if some of the axioms
introduced actually contradict other axioms. To prove that the axioms are
not contradictory, one would have to construct a model of the plane for
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ANCA MUSTATA
which all the axioms hold true, using other known mathematical objects
like numbers, vector spaces, etc.
Towards the end of the 19th century, David Hilbert began an immense
effort to construct a sound axiomatic basis for each area of mathematics.
His lectures at the university of Göttingen in 1898–1899, published under
the title Foundations of Geometry, proposed a larger set of axioms substituting the traditional axioms of Euclid. Hilbert proved that his axioms
are independent and non-contradictory (relying on algebra and coordinates
to construct a model of the plane satisfying his axioms). Since then, the
algebraic/analytic approach to Euclidian geometry has become dominant.
Time permitting, we will discuss Hilbert’s approach towards the end of the
course.
Independently and contemporaneously, a 19-year-old American student
named Robert Lee Moore published an equivalent set of axioms. Some
of the axioms coincide, while some of the axioms in Moore’s system are
theorems in Hilbert’s and vice-versa.
4. Direct consequences of the axioms
4.1. Angle measures. Define 1◦ as the 90-th part of a right angle, i.e. the
measure of an angle α such that 90 copies of α add up to a right angle. This
definition makes sense due to Euclid’s Postulate (4). Euclid doesn’t tell us
that such an angle α exists! But our intuition about the continuous nature
of the plane tells us that α exists.
Degrees are defined based on the notion of right angles (and the assumption that they are all equal), so if you try to define a right angle as being
90◦ , your definitions would be moving in circles. Similarly if you tried to
define supplements as summing up to 180◦ .
Proposition 4.1 (Opposite angles). Consider a line AB, a point O on it
in between A and B, and two points C and D in plane, on each side of the
line AB respectively. Then the points C, D and O are collinear if and only
[ = BOD.
\
if AOC
b
C
B
b
O
b
A
b
D
b
Proof. Here “If and only if” means that you have to prove two statements:
[ = BOD
\
a) If the points C, D and O are collinear then the angles AOC
are equal.
[ = BOD
\ then C, D and O are collinear.
b) If AOC
Prove a) by first noticing that supplements add up to 180◦ , and then that
[ and BOD
\ are supplements of the same angle BOC.
\ Prove b) by
both AOC
EUCLIDEAN GEOMETRY
9
contradiction. Assuming C, D and O are not collinear, extend the line CO
\ = BOE.
\
on the other side of AB by OE and then use a) to prove that BOD
Argue (using one of the common notions) that in this case the lines OE and
OD should coincide.
Remember Euclid’s Postulate 5): Given a point not on a given line, there
exists a unique line through that point parallel to the given line. In general,
if a straight line l intersects two other straight lines a and b, the sum of the
interior angles on the same side of l satisfies one of the following properties:
b
b
b
b
b
b
a
b
b
b
b
b
b
b
b
b
b
l
b
b
sum =
180◦
b
sum > 180◦
sum < 180◦
In the first case, we expect that the lines do not intersect. Two lines a and
b in the plane which do not intersect (no matter how far we extend them)
are parallel, denoted a k b.
Theorem 4.2 (Parallel lines I). Let AB and CD be two distinct lines
crossed by another line at the points P and Q like in the figure below. Then
\
\
AB k CD if and only if BP
Q+P
QD = 180◦ .
b
b
A
P
b
b
Q
C
b
b
B
D
Proof. The proof has two parts.
\
\
a) We assume that BP
Q+P
QD = 180◦ and prove AB k CD. Proof
by contradiction: Assume AB 6k CD. Then AB meets CD at a point
M , on one side of the line P Q, for example on the same side as B
and D.
b
N
b
A
b
b
B
b
b
C
P
b
Q
D
b
M
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ANCA MUSTATA
Then on the other side of the line P Q we can construct a point N
such that △M QP ≡ △N P Q. Indeed, it is sufficient to choose N on
the line AB such that |P N | = |QM |. Then
\
\
\
M
QP = 180◦ − M
PQ = N
P Q,
the last equality being due to the fact that M, B, P, A, N form a line.
\
\
Thus by placing the angle M
QP on top of the angle N
P Q so that Q
is placed on top of P and P on top of Q, then △M QP can fit exactly
on top of △N QP which means that the two triangles are congruent.
On the other hand, this implies that
\
\
\
\
N
QP = M
P Q = 180◦ − N
P Q = 180◦ − M
QP ,
so the points M, D, Q, C, P are collinear. Since M, B, P, A, N also
form a line, this would mean that the lines AB and CD are not
distinct, which contradicts the hypothesis. The contradiction is due
to our assumption that AB 6k CD. It follows that AB k CD.
\
\
b) We assume: AB k CD. We prove: BP
Q+P
QD = 180◦ . Proof by
\
\
contradiction: We can construct a line P E such that BP
Q+P
QE =
◦
180 , with D and E on the same side of line P Q. Then by Part a),
it follows that AB k QE. As we know AB k QD and Euclid’s 5th
Postulate states that through the point Q there should pass a unique
line parallel to AB, it follows that Q, E, D are collinear and hence
\
\
\
\
P
QD = P
QE. Thus BP
Q+P
QD = 180◦ as required.
Theorem 4.3 (Parallel lines II). The following statements are equivalent:
(p) AB k CD
\
\
(a1) BP
Q+P
QD = 180◦ : two interior consecutive angles add up to
◦
180 .
[
\
(a2) AP
Q+P
QC = 180◦ : two interior consecutive angles add up to
◦
180 .
\
[
(b1) P
QD = AP
Q: two alternate angles are equal.
\
\
(b2) BP
Q=P
QC: two alternate angles are equal.
[
\
(c1) EP
A=P
QC: two corresponding angles are equal.
\
\
(c2) EP B = P QD: two corresponding angles are equal.
\
\
(c3) QP
B=F
QD: two corresponding angles are equal.
\
[
(c4) F
QC = QP
A: two corresponding angles are equal.
b
A
P
E
B
b
b
b
C
Q
F
b
b
b
D
b
EUCLIDEAN GEOMETRY
11
Proof. Theorem 4.2 proves that (p), (a1) and (a2) are equivalent. To prove
that (a1) holds if and only if (b1) holds, note that both
[
\
\
\
AP
Q + QP
B = 180◦ and P
QD + QP
B = 180◦ .
Proof that (a) holds if and only if (c1) holds, and all the other equivalences,
can be proven in a similar way.
Theorem 4.4. The sum of all the interior angles of a triangle is 180◦ .
b
D
b
B
A
b
b
E
b
C
Proof. Consider △ABC. There exists a unique line DE passing through A
such that DE k BC, like in the diagram. Using the crossing lines AB and
AC, the previous theorem implies:
\ = BAD
\ and ACB
\ = CAE
[
ABC
(pairs of alternate angles). Then
\ + BAC
\ + ACB
\ = BAD
\ + BAC
\ + CAE,
[
ABC
\
= DAE,
= 180◦ .
5. Tessellations
A polygon is regular if all of its sides are equal and all of its angles are
equal. A tessellation or tiling of the plane is a collection of plane figures that
fills the plane with no overlaps and no gaps. The classic two-dimensional
picture of a beehive is a tiling made out of regular hexagons. This makes the
beehive into a sturdy construction, comfortable for the bees and suitable for
their communal life. But why don’t the bees construct their beehives out
of pentagonal, or octagonal shapes? In mathematical terms, we could pose
this question as follows:
Question 1. For which integer numbers n ≥ 3 does there exist a tiling of
the plane by identical regular n-sided polygons?
Idea for solution. To approach this problem, we assume that there exists
a tiling by identical regular n-sided polygons, and look at all the angles
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ANCA MUSTATA
around a vertex A of the tiling. Their sum is 360◦ , and they are all equal
to each other. It would be helpful if we knew the size of one such angle:
Question 2. What is the size of an interior angle of a regular polygon
with n sides?
For this, it would be helpful if we knew
Question 3. What is the sum of all the sizes of all interior angles of a
polygon with n sides?
Finally something we can start to answer right away:
Answer 3. Denote by S(n) the sum of all the sizes of all interior angles of
a polygon with n sides.
n angle
3
4
5
180◦
360◦
540◦
Indeed, S(3) = 180◦ is known; the interior of a quadrilateral can be split
into 2 triangles, and a pentagon into 3 triangles. We can thus prove
S(n) = (n − 2)180◦
(5.1)
by induction on n. For the induction step, we assume S(n) = (n − 2)180◦
and need to prove S(n + 1) = (n − 1)180◦ . For this, we split the interior
of a polygon of (n + 1) sides into one of n sides and a triangle. Thus
S(n + 1) = S(n) + 180◦ from our construction, and S(n) = (n − 2)180◦ from
our assumption. Putting these together we get S(n + 1) = (n − 1)180◦ . Now
we have
Question 2. What is the size of an interior angle of a regular polygon
with n sides?
Answer 2. An interior angle of a regular polygon with n sides measures
(n−2)
◦
n 180 :
sides angles
3
4
5
6
60◦
90◦
108◦
120◦
Question 1. For which integer numbers n ≥ 3 does there exist a tiling of
the plane by identical regular n-sided polygons?
Answer 1. We return now to our tiling by identical regular n-sided polygons, and assume that there are k angles around a vertex A of the tiling, for
an unknown integer k. Answer 2) above leads to the equation
k
2n
(n − 2)
180◦ = 360◦ , or, after simplifying, k =
.
n
n−2
EUCLIDEAN GEOMETRY
13
2n
is a whole number, n − 2 divides 2n. But n − 2 also divides
Since k = n−2
2(n − 2) = 2n − 4, and so n − 2 divides 4, i.e. n − 2 = 1, 2 or 4, so n = 3, 4
or 6.
Exercise. Find all pairs of integer numbers m > n ≥ 3 such that there
exists a tiling of the plane made only of regular polygons of n sides and m
sides. For each pair, make a sketch of a possible tiling.
m−2
Hint: Solve the equation k n−2
n + l m = 2 case by case. Note that in
5
this case k, l ≥ 1. Start with the case k = 1, n = 3. Then l m−2
m = 3 or,
5
equivalently, m−2
m = 3l . When l = 1, 3 or 4, the equation has no integer
solution m. When l = 2, we get m = 12. When l = 5 we get m = n = 3. If
l > 5 then m < 3 which contradicts our basic assumption about m.
Continue with the case k = 2, n = 3 etc.:
k n
l
1
2
3
4
1
2 12
2 6
2 4
1 6
2 8
3
3
3
3
4
m
6. Congruent triangles
Two triangles △ABC and △A′ B ′ C ′ are congruent if one can be superposed exactly on the other such that the point A coincides with A′ , the point
B with B ′ and the point C with C ′ . In this case we write
△ABC ≡ △A′ B ′ C ′ .
Corresponding parts of congruent triangles are congruent:
|AB| = |A′ B ′ |, Â = Â′ ,
|AC| = |A′ C ′ |, B̂ = B̂ ′ ,
|BC| = |B ′ C ′ |, Ĉ = Ĉ ′ .
Theorem 6.1 (Side-Angle-Side or SAS). If two triangles △ABC and △A′ B ′ C ′
satisfy
b
A
b
B
b
b

C B′
b
A′
b
C′

|AB| = |A′ B ′ | 
′
then: △ABC ≡ △A′ B ′ C ′
 = Â

|AC| = |A′ C ′ | 
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ANCA MUSTATA
Proof. Since  = Â′ , we can superpose the two angles such that the rays
AB and A′ B ′ coincide, and the rays AC and A′ C ′ coincide. Then on the
ray AB, we measure a length segment |AB| = |A′ B ′ |. This will place B ′
exactly in the same place as B. Similarly for C and C ′ .
Theorem 6.2 (Angle-Side-Angle or ASA). If two triangles △ABC and
△A′ B ′ C ′ satisfy
b
B
A
b
b
b
 = Â′
|AB| = |A′ B ′ |
B̂ = B̂ ′



A′
C B′
b
b
C′
then: △ABC ≡ △A′ B ′ C ′ .


Proof. Since |AB| = |A′ B ′ |, we can superpose the two segments such that
A and A′ coincide, and B and B ′ coincide. Then since  = Â′ , we can
superpose the ray A′ C ′ on AC. Since B̂ = B̂ ′ , we can superpose the ray
B ′ C ′ on BC. Thus C, the intersection of AC and BC, will coincide with
C ′ , the intersection of A′ C ′ and B ′ C ′ will coincide with .
As an application we have:
Theorem 6.3 (Isosceles triangle). Consider a triangle ABC. The following
two statements are equivalent:
(1) △ABC is isosceles with |AB| = |AC|.
(2) B̂ = Ĉ.
A
b
B
b
b
C
Proof.
a) We assume (1) △ABC is isosceles with |AB| = |AC|. We prove
(2)B̂ = Ĉ by comparing △ABC with its flip along the vertical axis
△ACB as follows:
|AB| = |AC|
\
\
ABC = ACB
|BC| = |CB|
Thus also B̂ = Ĉ.



SAS: △ABC ≡ △ACB


b) We assume (2)B̂ = Ĉ. We then prove (1) |AB| = |AC| by showing
that △ABC ≡ △ACB by (ASA).
EUCLIDEAN GEOMETRY
15
Theorem 6.4 (Side-Side-Side or SSS). If two triangles △ABC and △A′ B ′ C ′
satisfy
|AB| = |A′ B ′ | 
|AC| = |A′ C ′ | then: △ABC ≡ △A′ B ′ C ′ .

|BC| = |B ′ C ′ |

Proof. With no assumptions about angles, we cannot complete the superposing argument as before. We will then return to proof by contradiction.
Place △A′ B ′ C ′ on the same side of the line AB as △ABC, such that the
segment [A′ B ′ ] is superposed on [AB], i.e. such that B ′ coincides with B and
A′ with A, but assume that C 6= C ′ . This can happen in several cases:
a) If C ′ is in the interior of the triangle ABC.
b
b
A
b
C
C′
b
B
From the assumption SSS, the triangles ACC ′ and BCC ′ are isosceles with |AC| = |AC ′ | and |BC| = |BC ′ |, which by the previous
Theorem implies
′ C and BCC
′ C.
\′ = AC
\
\′ = BC
\
ACC
Adding up:
(6.1)
′ C + BC
′ C = 360◦ − AC
′ B.
\′ + BCC
\′ = AC
\
\
\
\ = ACC
ACB
′ B < 180◦ , so 360◦ −
\
\ < 180◦ and on the other hand AC
But ACB
′ B > 180◦ . Contradiction with the equality 6.1! Thus C ′ is not
\
AC
in the interior of the triangle ABC.
b) If C is in the interior of the triangle ABC ′ , we repeat the same proof
as in Case 1., after swapping C ′ and C.
′ C and BCC
′ C.
\′ = AC
\
\′ = BC
\
c) Like in Part a), ACC
16
ANCA MUSTATA
b
A
C′
b
C
b
b
B
′ B which is impossible,
\
\ = −AC
This time by subtracting, we get ACB
since one is positive and the other one negative.
7. Triangles in parallelograms
A quadrilateral whose opposite sides are parallel is a parallelogram.
Theorem 7.1 (Parallelogram). Let ABCD be a quadrilateral. The following statements are equivalent:
a) ABCD is a parallelogram.
b) AB k CD and |AB| = |CD|.
c) The diagonals AC and BD intersect at their midpoint.
d) |AB| = |CD| and |BC| = |AD|.
B
b
b
b
A
b
C
E
b
D
Proof. The four statements a)-d) are equivalent if each of them implies any
other among them. Thus it seems that we would have to prove a total of
12 implications! (check that 12 is the correct number). However, we can
EUCLIDEAN GEOMETRY
17
considerably shorten our work by following a diagram like this one:
a)
Part a)
Part d)
b)
Part b)
d)
Part c)
c)
Note that we can connect any two points in the diagram by a sequence of
arrows. Other strategies (arrow diagrams) can be designed for the same
problem, but this is one of the shortest with just 4 implications.
a) We assume a) and prove b). We only need to prove |AB| = |CD|.
\ = DBC.
\
Note that BC k AD so BDA
\ = CDB
\
ABD
|BD| = |DB|
\ = DBC
\
BDA
So |AB| = |CD|.



ASA: △BDA ≡ △DBC


b) We assume b) and prove c). Let E denote the intersection point
\ = CDB.
\
of the diagonals. Alternate angles in ABkCD give ABD
\ = DCA.
\
Alternate angles in ABkCD give BAC
\ = ECD
\
EAB
|AB| = |CD|
\ = CDE
\
ABE



ASA: △ABE ≡ △CDE


so |AE| = |CE| and |BE| = |DE|.
\ = DEC.
\
c) We assume c) and prove d). By opposite angles, BEA


|AE| = |CE| 
\
\
AEB = CED  SAS: △AEB ≡ △CED
|EB| = |ED| 
so |AB| = |CD|. Similarly, we can prove △CBE ≡ △ADE and
hence |CB| = |AD|.
d) We assume d) and prove a).
|AB| = |CD| 
|BC| = |DA| SSS: △ABC ≡ △CDA

|AC| = |CA|

\ = DCA
\ and ACB
\ = CAD.
\ So AB k CD and BC k AD by
So BAC
the theorem on parallel lines.
18
ANCA MUSTATA
7.1. Special parallelograms. A quadrilateral with four right angles is a
rectangle. A rectangle is a parallelogram by the Theorem on parallel lines
(4.3).
Theorem 7.2 (Rectangle). Let ABCD be a quadrilateral. The following
statements are equivalent:
a) ABCD is a rectangle.
b) The diagonals AC and BD intersect at their midpoint and |AC| =
|BD|.
Proof.
a) Assume a) and prove b). By the theorem on parallelograms, |AB| =
|CD|.

|BA| = |CD| 

\ = CDA
\ SAS: △BAD ≡ △CDA
BAD

|AD| = |DA| 
So |AC| = |BD|.
b) Assume b) and prove a). We first note that ABCD is a parallelogram
by the Theorem on parallelograms (C), since the diagonals AC and
BD intersect at their midpoint. By the theorem on parallelograms,
|AB| = |CD|.
|AB| = |DC| 
|BD| = |CA| SSS: △ABD ≡ △DCA

|AD| = |DA|

\ = CDA
\ and, since the sum of the two angles is 180◦ by the
So BAD
\ = CDA
\ = 90◦ . Then
Theorem on parallel lines, it follows that BAD
\ + ABC
\ = 180◦ and
again, by the Theorem on parallel lines, BAD
◦
◦
\ + DCB
\ = 180 , so ABC
\ = DCB
\ = 90 .
CDA
A quadrilateral all of whose sides are equal is a rhombus. Any rhombus
is a parallelogram by the theorem on parallelograms d).
Theorem 7.3 (Rhombus). Let ABCD be a quadrilateral. The following
statements are equivalent:
a) ABCD is a rhombus.
b) The diagonals AC and BD intersect at their midpoint and AC ⊥
BD.
Proof.
a) We assume a) and prove b). Triangles △BAC and △DAC are
isosceles with |AB| = |BC| = |CD| = |DA| so
\
\ = BCA
\ = DAC
\ = DCA
\ = 1 BAD.
(7.1)
BAC
2
EUCLIDEAN GEOMETRY
19
Similarly, △ABD and △CBD are isosceles with |AB| = |BC| =
|CD| = |DA| so
(7.2)
\ = ADB
\ = CBD
\ = CDB
\ = 1 ADC.
\
ABD
2
On the other hand, AB k CD so
\ + ADC
\ = 180◦ .
BAD
This, together with equations (7.1) and (7.2), implies
\ + ABD
\ = 90◦ .
BAC
b) We assume b) and prove a). Let E be the intersection point of AC
and BD.
|AE| = |CE|
\
\
AEB = CEB
|EB| = |EB|



SAS: △AEB ≡ △CEB


So |AB| = |BC|. By the Theorem on parallelograms, all sides are
equal.
8. The Pythagorean theorem
In this section we will employ the notion of area of a bounded plane figure.
The basic principles defining the concept of area are the following:
a) The area of a rectangle ABCD is equal to the product of its length
and height: |AB| · |AD|.
b) Two congruent figures have the same area.
Lemma 8.1. The area of a triangle ABC with  = 90◦ is 12 |AB| · |AC|.
Proof. We complete the triangle ABC to a rectangle ABDC. By the Theorem on rectangles above, △ABC ≡ △DCB, and as such, each of the
triangles has an area equal to half the area of the rectangle.
Theorem 8.2 (The Pythagorean theorem). Let ABC be a triangle with
 = 90◦ . Then
|AB|2 + |AC|2 = |BC|2 .
Proof. On the ray AB, extend the segment [AB] by a segment [BM ] such
that |BM | = |AC|.
20
ANCA MUSTATA
N
b
b
D
P
b
C
b
E
b
b
A
b
B
b
M
On the ray AC, extend the segment [AC] by a segment [CN ] such that
|CN | = |AB|. Construct the square M AN P and let D ∈ [N P ] and E ∈
[P M ] be such that
|M B| = |AC| = |N D| = |P E| and |M E| = |AB| = |N C| = |P D|.
These relations, together with M̂ = Â = N̂ = P̂ = 90◦ , imply
△M EB ≡ △ABC ≡ △N CD ≡ △P DE.
From here:
\
\=N
\
\
\
M
EB = ABC
CD = P
DE, M
BE
\=N
\
\
= ACB
DC = P
ED,
|BE| = |CB| = |DC| = |ED|.
\ = 180◦ − ABC
\−M
\
\ − ACB
\ = 90◦ , and
From here, CBE
BE = 180◦ − ABC
similarly
\ = BCD
\ = CDE
\ = DEB
\ = 90◦ .
EBC
Thus BCDE is a square, since all its angles are right angles and all its sides
are equal. We can now compare areas:
Area AN P M = Area BCDE + 4 Area ABC,
1
so (|AB| + |AC|)2 = |AM |2 = |BC|2 + 4 · |AB| · |AC|,
2
so |AB|2 + |AC|2 = |BC|2 .
Corollary 8.3 (The congruence case). Two right angle triangles △ABC
and △A′ B ′ C ′ having two pairs of sides of equal lengths |AB| = |A′ B ′ | and
|AC| = |A′ C ′ | are congruent triangles.
Proof. Applying Pythagoras’ theorem to both triangles will result in |BC| =
|B ′ C ′ | as well so we can use SAS or SSS.
EUCLIDEAN GEOMETRY
21
9. Special lines in a triangle
In this section we will use congruence of triangles to study five types of
important lines in a triangle: the angle bisectors, the perpendicular bisectors,
the altitudes and the medians, as well as the midlines. We will prove that
the three angle bisectors of a triangle intersect at a unique point. Similarly
for the three perpendicular bisectors; for the three altitudes and for the three
medians. We will also study the defining properties of the points forming
these lines.
\ is the line AD
9.1. Angle bisector. The angle bisector of an angle BAC
\
such that D is a point in the interior of the angle and BAC
\ = DAC.
\
BAD
C
b
b
A
b
D
B
b
9.2. Constructing angle bisectors. Draw an arc of circle centered at A,
and let the intersection points with the rays AB and AC be E and F ,
respectively. Two arcs of circles centered at E and F respectively, and of
the same radius, will intersect at a point D.
F
b
b
b
A
b
b
D
b
E
\ Indeed,
Then AD is the angle bisector of BAC.
|AE| = |AF | 
|ED| = |F D| SSS: △AED ≡ △AF D

|AD| = |AD|

\ = DAF
\.
So EAD
There is an alternative way of characterizing the bisector, involving the
distance from a point to a line.
The distance from a point D to a line AB not containing the point is
the length |DM |, where DM ⊥ AB and M ∈ AB. Note that this is the
22
ANCA MUSTATA
same as the shortest path from D to a point on AB, due to the Pythgoraean
theorem. A geometric locus is a set of points in the plane all of which satisfy
a given property.
Theorem 9.1 (The angle bisector as a geometric locus). Consider an an\ and a point D in its interior. The following two statements are
gle BAC
equivalent:
\
a) AD is the angle bisector of BAC.
b) The point D is equally distanced from AB and AC.
In other words, the angle bisector is the geometric locus of all the points
in the interior of an angle equally distanced from the sides of the angle.
B
b
M
b
D
b
A
b
b
N
C
b
Proof. Consider the points M ∈ AB and N ∈ AC such that DM ⊥ AB and
DN ⊥ AC.
a) Assume a) and prove b). By right angles,
\ = 90◦ − DAM
\ = 90◦ − DAN
\ = ADN
\
ADM
so
\
\
M
AD = N
AD
|AD| = |AD|
\ = ADN
\
ADM



ASA: △ADM ≡ △ADN


So the distances |M D| and |N D| from D to AB and AC, respectively,
are equal.
b) Assume b) and prove a). By the Pythagorean Theorem,
|AM | =
q
|AD|2
−
|M D|2
=
q
|AD|2 − |N D|2 = |AN |.
|M D| = |N D| 
|DA| = |DA|
SSS: △M DA ≡ △N DA

|M A| = |N A|

\
\ . Thus AD is the angle bisector of M
\
\
So M
AD = DAN
AN = BAC.
This universal property of a bisector is helpful in proving an important
property of a triangle:
EUCLIDEAN GEOMETRY
23
Theorem 9.2 (The incentre of a triangle). All of the angle bisectors of the
interior angles in a triangle ABC intersect at a point I, the incentre of the
triangle.
b
A
b
M
NE
b
b
I
b
B
b
b
b
b
C
P D
\ and
Proof. We first notice that the angle bisectors AD and BE of BAC
\ respectively, must intersect at a point. Indeed, if they were parallel,
ABC
\ + EBA
\ = 180◦ . But DAB
\=
then by the theorem on parallel lines, DAB
1\
1\
◦
\
\
\
2 BAC and EBA = 2 ABC, and so we would have BAC + ABC = 360 ,
which is absurd: they are interior angles in △ABC and so their sum should
be 180◦ .
We will denote the point of intersection of AD and BE by I. It remains
\ Let |IM |, |IN |, and |IP |
to show that CI is the angle bisector of ACB.
be the distances from I to AB, AC and BC respectively, with M ∈ AB,
N ∈ AC, P ∈ BC.
\ by the universal property of angle
Since AD is the angle bisector of BAC,
bisectors we have |IM | = |IN |.
\ by the universal property of angle
Since BE is the angle bisector of ABC,
bisectors we have |IM | = |IP |
Thus |IN | = |IP | and so, by the universal property of angle bisectors, CI
\
is the angle bisector of ACB.
In the proof above, we’ve seen that |IM | = |IN | = |IP |, and so, there
exists a circle with centre at I and radius IM , the incircle of △ABC. Because IM ⊥ AB, IN ⊥ AC and IP ⊥ BC, the incircle intersects the sides
of △ABC at the points M, N, P only. The incircle is tangent to the sides of
△ABC, i.e. touches each side once. Indeed, if the incircle intersected AB
at two points M and M ′ , then △IM M ′ would be isosceles with the angles
at M and M ′ both equal to 90◦ , which is impossible.
9.3. The perpendicular bisector. The perpendicular bisector of a segment [BC] is the line perpendicular on [BC] and passing through its midpoint. To construct the perpendicular bisector, draw two circles centered at
B and C respectively, and of the same radius, which intersect at two points
S and T . Then the line ST is the perpendicular bisector of [BC]. Indeed,
By construction, |BS| = |CS| = |BT | = |CT | and so BSCT is a rhombus.
24
ANCA MUSTATA
By the theorem on rhombi, we know that ST ⊥ BC and that ST and BC
intersect at their midpoints.
There is an alternative way of characterizing the perpendicular bisector.
Theorem 9.3 (The perpendicular bisector as a geometric locus). Consider
a segment [BC] and a point S not on it. The following two statements are
equivalent:
a) S lies on the perpendicular bisector of [BC].
b) |SB| = |SC|.
In other words, the perpendicular bisector of a segment is the geometric
locus of all the points equidistant from the vertices of the segment.
Proof. Consider the point M ∈ BC such that SM ⊥ BC.
a) Assume a) and prove b). By assumption, M is the midpoint of [BC]
and SM ⊥ BC, thus

|BM | = |CM | 

\
\
BM
S = CM
S  SAS: △BM S ≡ △CM S
|M S| = |M S| 
So |SB| = |SC|.
◦
\
\
b) Assume b) and prove a). By construction,
SM
B = SM
p C = 90 . By
p
2
2
2
Pythagoraean Theorem, |BM | = |SB| − |M S| = |SC| − |M S|2 =
|CM |. Thus SM is the perpendicular bisector of [BC].
This universal property of a perpendicular bisector is helpful in proving
an important property of a triangle:
Theorem 9.4 (The circumcentre of a triangle). All the perpendicular bisectors of the sides in a triangle ABC intersect at a point O, the circumcentre
of the triangle, because there exists a circle, the circumcircle of the triangle,
of centre O and containing the vertices of the triangle ABC.
b
A
b
b
O
b
B
b
b
b
C
Proof. We first notice that the perpendicular bisectors of [BC] and [AB]
must intersect at a point, which we will call O. Indeed, if they were parallel,
then by drawing a common parallel to both through B and applying the
\ = 180◦ , which is absurd.
theorem on parallel lines, we would get ABC
EUCLIDEAN GEOMETRY
25
It remains to show that O lies on the perpendicular bisector of [AC]. Since
O lies on the perpendicular bisector of [BC], by the universal property of
perpendicular bisectors we have |OB| = |OC|.
Since O lies on the perpendicular bisector of [BA], by the universal property of perpendicular bisectors we have |OB| = |OA|.
Thus |OA| = |OC| and so, by the universal property of perpendicular
bisectors, O lies on the perpendicular bisector of [AC].
Since |OA| = |OB| = |OC|, the point O is the centre of a circle which
contains all vertices of the triangle ABC.
Example 9.5. The circumcentre of a right angled triangle is always the
midpoint of the hypothenuse.
Indeed, we can complete △ABC with ∡A = 90◦ to a rectangle ABA′ C.
From the theorem of rectangles, we know that the diagonals AA′ and BC
are equal and intersect at their midpoint O. Thus O is equally distanced
from A, B, C, A′ .
9.4. Altitudes. The altitude from the vertex A of a triangle ABC is the
line through A perpendicular on BC.
Example 9.6. The lines AD, EF and HK are altitudes in the three distinct
triangles below:
b
E
A
b
b
B D
b
b
H
b
C F
b
b
G
b
K
b
I
b
J
Theorem 9.7 (Orthocentre). All of the altitudes of a triangle intersect at
a point H, the orthocenter of the triangle.
b
C′
P
B
b
B′
A
b
b
b
N
b
b
b
M
b
C
A′
Proof. Through each of the vertices of the triangle ABC we draw a parallel
to the opposite side. By intersecting these lines we get another triangle
A′ B ′ C ′ , such that A ∈ B ′ C ′ , B ∈ A′ C ′ , C ∈ A′ B ′ .
26
ANCA MUSTATA
The proof is based on the observation that the altitudes in the triangle
ABC are perpendicular bisectors in triangle A′ B ′ C ′ . Since we have shown
that the perpendicular bisectors of a triangle are concurrent, it will follow
that the altitudes in the triangle ABC are concurrent at a point H.
Indeed, ABCB ′ and ACBC ′ are parallelograms, so by the theorem on
parallelograms, |AB ′ | = |BC| and |BC| = |AC ′ |, which imply |AB ′ | =
|AC ′ |. Moreover, since BC k B ′ C ′ , it follows that the altitude AM of
the triangle ABC is also perpendicular to [B ′ C ′ ], and passing through its
midpoint A. Thus AM is the perpendicular bisector of [B ′ C ′ ]. Similarly,
the altitude BN is the perpendicular bisector of [A′ C ′ ] and the altitude CP
is the perpendicular bisector of [A′ B ′ ].
As in the case of angle and perpendicular bisectors, it would be nice if
we could express an altitude as a geometric locus, i.e. if we could decide
whether a point H is on the altitude from A to BC based on measurements
involving only H, A, B and C. Given a protractor, we could simply check
whether the angle between the lines AH and BC is 90◦ . Given only a ruler,
we could make use of Pythagora’s theorem to check whether the same angle
is 90◦ . Since our measurement should involve only segments made by H, A,
B and C, we will come up with a slightly complicated condition obtained
by applying Pythagoras’ theorem a number of times and cancelling some
irrelevant terms.
Lemma 9.8. Consider △ABC. Take a point D on the line BC.
AD ⊥ BC just when |AB|2 − |AC|2 = |DB|2 − |DC|2
Proof.
a) Assume AD ⊥ BC. Prove the relation above. Indeed, D̂ = 90◦ . By
Pythagoras’ theorem in △ADB and △ADC we have:
|AD|2 + |DB|2 = |AB|2 ,
|AD|2 + |DC|2 = |AC|2 .
After subtracting the two equations term by term and canceling out
|AD|:
|DB|2 − |DC|2 = |AB|2 − |AC|2 .
b) Assume |DB|2 − |DC|2 = |AB|2 − |AC|2 . We’ll prove AD ⊥ BC.
Indeed, assume D′ is the point on BC such that AD′ ⊥ BC. Then
by part I,
|D′ B|2 − |D′ C|2 = |AB|2 − |AC|2 , while
|DB|2 − |DC|2 = |AB|2 − |AC|2
by assumption. Hence |D′ B|2 − |D′ C|2 = |DB|2 − |DC|2 , or equivalently,
(9.1)
(|D′ B| − |D′ C|)(|D′ B| + |D′ C|) = (|DB| − |DC|)(|DB| + |DC|).
EUCLIDEAN GEOMETRY
27
But |D′ B|+|D′ C| = |DB|+|DC| = |BC| if both D and D′ are inside
the segment [BC], (which happens when △ABC is acute-angled), or
|D′ B| − |D′ C| = |DB| − |DC| = ±|BC| if D and D′ are outside the
segment [BC], both on the same side of the vertices B and C (which
is the case when B̂ > 90◦ or Ĉ > 90◦ ).
After cancelling out the equal factors in equation (9.1), we have both
|D′ B| + |D′ C| = |DB| + |DC| and
|D′ B| − |D′ C| = |DB| − |DC|.
which added/subtracted yield |D′ B| = |DB| and |D′ C| = |DC|,
hence D = D′ .
We note that it would be impossible for one of D, D′ to be inside the
segment [BC] and the other outside. Indeed, if for example D′ is outside and D inside, then |D′ B| − |D′ C| = |DB| + |DC| = |BC|. After
canceling these factors in equation (9.1), we’d get |D′ B| + |D′ C| =
|DB| − |DC| which is impossible as the positions of D′ and D imply
|D′ B| + |D′ C| ≥ |D′ B| > |BC| > |DB| > |DB| − |DC|.
In the right-angled triangle △ADB with D̂ = 90◦ we define
cos B :=
|AD|
|BD|
and sin B :=
.
|AB|
|AB|
Note: cos B and sin B thus defined seems to depend on the choice of the
triangle △ABD. We will prove later that they depend in fact only on the
measure of the angle B̂.
Corollary 9.9 (The cosine formula). Consider △ABC with the side lengths
denoted by |AB| = c, |AC| = b and |BC| = a. Let D be the point on BC
such that AD ⊥ BC. Then
cos B =
a2 + c2 − b2
2ac
Proof. Consider the case when △ABC is acute angled. From Lemma 9.8 we
have |DB|2 − |DC|2 = |AB|2 − |AC|2 = c2 − b2 . But |DC| = |BC| − |DB| =
a − |DB|. Substituting this in the equation above we get: |DB|2 − (a −
2
2
2
|DB|)2 = c2 − b2 .. Solving for |DB| we get |DB| = a +c2a−b .
Corollary 9.10 (Heron’s formula for the area of a triangle). Consider
△ABC with the side lengths denoted by |AB| = c, |AC| = b and |BC| = a.
Then
Area ABC =
q
p(p − a)(p − b)(p − c)
where p = 21 (a + b + c) is the semiperimeter of △ABC.
28
ANCA MUSTATA
Proof. Continuing with the calculations from the previous proof, we apply
2
2
2
Pythagoras’ theorem in △ADB with |AB| = c and |DB| = a +b2a−c to get
|AD| =
=
q
s
|AB|2 − |DB|2 ,
c2 −
(a2 + c2 − b2 )2
,
4a2
s
4a2 c2 − (a2 + c2 − b2 )2
,
4a2
p
(2ac − a2 − c2 + b2 )(2ac + a2 + c2 − b2 )
=
,
2a
p
(b2 − (a − c)2 )((a + c)2 − b2 )
=
,
2a
p
(b + a − c)(b − a + c)(a + c − b)(a + c + b)
,
=
2a
p
4 p(p − a)(p − b)(p − c)
=
2a
As |AD| is the height in △ABC with basis |BC| = a, we get the formula
for the area as above.
=
Theorem 9.11 (Altitude as a geometric locus). Let △ABC be a triangle
and H a point in the plane. Then H is on the altitude from A to BC just
when
AH ⊥ BC which occurs just when |AB|2 − |AC|2 = |HB|2 − |HC|2
Proof. Let D be the intersection of the lines AH and BC.
a) Assume AH ⊥ BC. We will prove the formula above. Indeed,
applying Lemma 9.8 to AD ⊥ BC and then HD ⊥ BC we get
|AB|2 − |AC|2 = |DB|2 − |DC|2 = |HB|2 − |HC|2
b) Assume |AB|2 − |AC|2 = |HB|2 − |HC|2 . We will prove AH ⊥ BC.
Let D, D′ be points on BC such that AD ⊥ BC and then HD′ ⊥ BC.
Thus by Lemma 9.8,
|AB|2 − |AC|2 = |DB|2 − |DC|2 and |HB|2 − |HC|2 = |D′ B|2 − |D′ C|2 .
We assumed |AB|2 −|AC|2 = |HB|2 −|HC|2 , and so |DB|2 −|DC|2 =
|D′ B|2 − |D′ C|2 which as before implies D = D′ . Thus the lines
AD ⊥ BC and HD′ ⊥ BC have a common point D = D′ . But only
one perpendicular to BC can be constructed from the point D and
hence A, D and H must be collinear, AH ⊥ BC.
An alternate proof for the Orthocenter Theorem: Note that two altitudes
BE and CF must intersect at a point. Indeed, assume BE k CF . Then
BE ⊥ AC would imply CF ⊥ AC whereas from definition CF ⊥ AB. But
EUCLIDEAN GEOMETRY
29
then the triangle formed by the lines AB, AC and CF would have two right
angles, which is impossible.
We denote by H the intersection of the two altitudes BE and CF . We
will prove that H is also on a point on the altitude from A. Indeed, we can
apply the Altitude as geometric locus Theorem:
BH ⊥ AC just when |BA|2 − |BC|2 = |HA|2 − |HC|2
CH ⊥ AB just when |CB|2 − |CA|2 = |HB|2 − |HA|2 .
Subtract: |BA|2 − |CA|2 = |HB|2 − |HC|2 ,
which implies AH ⊥ BC.
9.5. Median. The median from the vertex A of a triangle ABC is the line
joining A with the midpoint of [BC]. The midpoint C ′ of the segment AB is
joined to the midpoint B ′ of AC by the midline B ′ C ′ of the triangle ABC.
Proposition 9.12 (Midlines). Let ABC be a triangle and consider the midpoint M of the segment AB and the midpoint N of AC.
a) M N kBC and |M N | = |BC|/2.
b) Let G be the point of intersection of BN and CM . Then |BG| =
2|GN | = 32 |BN | and |CG| = 2|GM | = 23 |CM |.
Proof. a): Extend M N by M P of equal length. By the Theorem on parallelograms (D), AM CP is a parallelogram. This implies that CP kAB
and |CP | = |AM | = |M B|. Thus by the Theorem on parallelograms
(B), BM P C is a parallelogram too, which implies M P kBC and 2|M N | =
|M P | = |BC|. This proves a).
b
b
A
M
b
N
b
P
G
b
b
B
b
R
b
S
b
C
b) Draw the midpoints R and S of the segments BG and CG, respectively.
Then N S is a midline in △CGA and so by a),
1
N SkAG and |N S| = |AG|.
2
30
ANCA MUSTATA
Similarly, M R is a midline in △BGA and so by a),
1
M RkAG and |M R| = |AG|.
2
From the previous two observations, N SkM R and |N S| = |M R|. By the
Theorem on parallelograms (B), M N SR is a parallelogram, and so the diagonals intersect each other at midpoints. Thus |M G| = |GS| = |SC| =
1/2|CG| (because S was constructed as the midpoint of the segment CG),
and similarly |N G| = |GR| = |RG| = 1/2|BG|.
Corollary 9.13. Take a triangle △ABC and split each side in half, drawing all 3 midlines. Then these midlines split the triangle into 4 congruent
triangles.
b
b
b
b
b
b
Proof. Each side of the triangle bound by the midlines is parallel to and
half the size of the obvious side of △ABC, so all four small triangles are
congguent by SSS.
Lemma 9.14. If △ABC and △A′ B ′ C ′ have the same angles at B and B ′ ,
and at C and C ′ , and if |BC| = |B ′ C ′ |, then every side of △A′ B ′ C ′ is double
the length of the corresponding side of △ABC.
Proof. Chop up △A′ B ′ C ′ by midlines as above, to find 4 triangles, all congruent to △ABC by ASA.
Theorem 9.15 (The centroid of a triangle). All of the medians of a triangle
intersect at a point G, the centroid of the triangle.
Proof. Let Q be the midpoint of [BC]. With the notations from the Proposition on Midlines, AQ, BN and CM are the median in the triangle ABC,
and G is the point of intersection of BN and CM . Assume that BN and
AQ intersect at another point G′ . We apply the Proposition on Midlines, b)
in two cases:
• to G as the point of intersection of BN and CM so |BG| = 23 |BN |.
• to G′ as the point of intersection of BN and AQ so |BG′ | = 23 |BN |.
From here it follows that G = G′ , since |BG| = |BG′ | and G, G′ are both
interior points of [BN ]. Thus AQ, BN and CM all intersect at G.
EUCLIDEAN GEOMETRY
31
Theorem 9.16 (Median as a geometric locus). A point G in the interior
of a triangle △ABC is on one of its medians AM if it forms with the sides
AB and AC triangles of equal areas:
G ∈ AM is a median just when Area ABG = Area ACG.
Proof.
a) Assume M is the midpoint of segment BC and that G is a point on
the median AM . Then
Area ABM = Area ACM
as the triangles have equal bases and the same height. Similarly,
Area GBM = Area GCM.
Subtracting the two equations above yields
Area ABG = Area ACG.
b) Assume Area ABG = Area ACG. As the two triangles have a common side AG, it follows that their heights |BE| and |CF | are equal.
b
A
b
G
90◦E
b
B
b
b
M
b
b
C
F
90◦
Then △BEM ≡ △CF M by AAS, as they have:
• |BE| = |CF |
\ = CF
\
• BEM
M = 90◦
\
\
• BM
E = CM
F as opposite angles.
Hence, |BM | = |CM | and so AM is median.
32
ANCA MUSTATA
An alternate proof for the theorem of the centroid:
Proof. Two medians BN and CP of △ABC will always intersect at a point
G, as they are both in the interior of △ABC. We will use the characterization of medians as geometric locus to prove that G is also a point on the
median AM . Indeed,
G ∈ BN = median just when Area BAG = Area BCG,
G ∈ CP = median just when Area BCG = Area CAG.
Hence Area BAG = Area CAG, and so G is also a point on the median
AM .
Theorem 9.17 (Euler’s line). The orthocentre H, circumcentre O, and
centroid G of any triangle △ABC are collinear and satisfy |HG| = 2|HO|.
A
b
b
b
b
H
b
B
b
G
O
b
b
M
b
b
C
A′
Proof.
• Let M be the midpoint of the segment BC and let AA′ be the diameter
of the circumcircle of △ABC. Then A′ B ⊥ AB and CH ⊥ AB so
A′ B k CH. As well, and A′ C ⊥ AC and BH ⊥ AC, so A′ C k BH
and so BHCA′ is a parallelogram.
• Hence M is the midpoint of segment [HA′ ].
• The point G is on one of the medians exactly in the right place for the
centroid of △AHA′ , cutting the median in ratio 2 : 1.
• Hence G is the centroid of △AHA′ . As such, G is on the median HO
and |HG| = 2|HO| by the Theorem of the Centroid.
EUCLIDEAN GEOMETRY
33
10. Circles
We will denote by Cr O a circle of center O and radius r.
10.1. Secants and tangents. A secant is a line which intersects the circle
at two points.
Lemma 10.1. Let Cr O be a circle of centre O and radius r and let A, B be
two points on the circle. Then the perpendicular from O to AB intersects
the segment [AB] at its midpoint P , and
1
|OP |2 = r2 − |AB|2 .
4
Proof. Let P denote the point of intersection of AB with the perpendicular
from O to AB. Then by applying Pythagora’s theorem in the right angled
triangles △OP A and △OP B we get
|AP |2 = |OA|2 − |OP |2 = |OB|2 − |OP |2 = |BP |2 .
As |AP | = 12 |AB|, the equation above can be rearranged like in the conclusion of the lemma.
A tangent is a line which intersects the circle at exactly one point. We will
think of the tangent as the limit of a sequence of secants passing through a
fixed point M , and moving gradually further away from the center O of the
circle.
Let M be a point outside a line d, and let P be a point on d such that
M P ⊥ d. We say that P is the projection of M on the line d.
Theorem 10.2. Let M be a point not situated on a circle C of center O, and
let M P be tangent to C, where P is a point on the circle. Then OP ⊥ M P .
Proof. We consider a secant line AB passing through M . Keeping M as a
pivot, we move the line AB towards the outside of the circle. Let P denote
the projection of O on the line AB, so that OP ⊥ M P . Then by the previous
lemma,
1
|OP |2 = r2 − |AB|2 .
4
The line through M , A, B and P is in a tangent position to the circle
precisely when |AB| = 0, and so |OP | = r. In other words, P is on the
circle and OP ⊥ M P .
Conversely, we have:
Theorem 10.3. Let M be a point not situated on a circle C with center O,
and let P is a point on the circle. Assume OP ⊥ M P . Then M P is tangent
to the circle C .
Proof. We want to show that M P is tangent to the circle C , i.e., by definition, that P is the unique point of intersection of M P with C . Uniqueness
is most often proven by contradiction, so we will assume that there exists
34
ANCA MUSTATA
another point of intersection P ′ . Then since P and P ′ are both on the
circle, the triangle OP P ′ is isosceles and as such, Pˆ′ = P̂ = 90◦ (since
OP ⊥ M P ). But the angles in △OP P ′ can’t sum up to more than 180◦ –
thus our assumption of the existence of P ′ must have been wrong.
Lemma 10.4. Let M be a point not situated on a circle C , and let M P
and M P ′ be tangents to C , where P and P ′ are points on the circle. Then
|M P | = |M P ′ |.
Proof. This follows by applying the Pythagorean theorem in triangles OM P
and OM P ′ .
Lemma 10.5. Consider two circles C1 = Cr1 O1 and C2 = Cr2 O2 with
common tangents M P and M ′ P ′ , where M, M ′ are points on C1 and P, P ′
are points on C2 . Then
|M P | = |M ′ P ′ | =
q
|O1 O2 |2 − (r1 − r2 )2 .
Hint: Draw O1 A ⊥ O2 P and O1 A ⊥ O2 P ′ and apply the Pythagorean
theorem in triangles O1 O2 A and O1 O2 A′ .
Theorem 10.6. Let C1 = Cr1 O1 and C2 = Cr2 O2 be two circles intersecting
at two points A and B. Then O1 O2 ⊥ AB and O1 O2 passes through the
midpoint of the segment [AB].
\
\
Proof. △AO1 O2 ≡ △BO1 O2 (case SSS) just when AO
2 O1 = BO2 O1 just
when O1 O2 is angle bisector in the isosceles triangle O2 AB with |O2 A| =
|O2 B| just when O1 O2 ⊥ AB and O1 O2 passes through the midpoint of
the segment [AB] (as O1 O2 must also be perpendicular bisector in triangle
O2 AB).
Two circles are tangent to each other if they intersect at only one point.
Just like with a circle and a line, we consider tangency of two circles as the
limit position of a sequence of secant circles. Hence the following theorem.
Theorem 10.7. Let Cr1 O1 and Cr2 O2 be two circles tangent to each other
at the point P . Then O1 , O2 and P are collinear.
Proof. Consider a sequence of points On on the line O1 O2 , approaching
O1 , but chosen so that the circles Cn = Cr1 On intersect the circle C2 =
Cr2 O2 at two points An and Bn . By the previous theorem, the line On O2
passes through the midpoint Pn of the segment [An Bn ]. On the other hand,
limn→∞ On = O1 implies limn→∞ An = P and limn→∞ Bn = P , and so
the same is true for midpoints: limn→∞ Pn = P (exercise in coordinate
calculations, using the fact that all circles Cn have radius r1 ). As Pn ∈ O1 O2
for all n, it follows that P ∈ O1 O2 .
EUCLIDEAN GEOMETRY
35
>
\
10.2. Arcs and angles. In a circle Cr O, the arc AB is the angle AOB.
Lemma 10.8 (Angle on a circle). Let A, B, C be three points on a circle.
>
Let BC denote the arc which does not contain A. Then
>
\
BC
BOC
\=
=
.
BAC
2
2
Proof. Let D be the point on the circle such that A, O and D are collinear.
Then
\ = 2OAB,
\ as exterior angle of the isosceles triangle OAB, and
• BOD
\ = 2OAC,
[ as exterior angle of the isosceles triangle OAC.
• COD
\ we add the equations above term by
If O is in the interior of the angle BAC,
\ we subtract the equations above
term, and O is outside of the angle BAC,
\ = 2BAC.
\
term by term. In both cases, we obtain BOD
Corollary 10.9 (Internal and external angles). Let C be a circle of center
O. Let A, B, C, D be points on C and P the intersection point of AB and
CD, which we suppose lies in the interior of C .
C
b
B
b
P
Q
b
b
b
D
b
A
Let Q be the intersection point of the lines AD and BC. Then
>
1 > >
[
[ = 1 (>
AP
C = (AC + BD) and AQC
AC − BD).
2
2
[
Proof. AP
C is exterior angle for △P BC so
[
\ + DCB,
\
AP
C = ABC
\
\
=P
BC + P
CB,
1 > >
= (AC + BD)
2
\ is exterior angle for △QBA so AQB
\ = ABC
\−
(angles on the circle). ABC
>
>
\ = 1 (AC − BD).
BAD
2
Lemma 10.10. Let A, B, C be three points on a circle. Also, let BD be
tangent to the circle, with D on the same side of AB as C. Then
\ = DBC.
\
BAC
The proof is left as an exercise.
A quadrilateral ABCD is cyclic if all its vertices are on a circle.
36
ANCA MUSTATA
Lemma 10.11 (Isosceles trapezoid in a circle). Let ABCD be a cyclic
quadrilateral. The following are equivalent:
a) |AD| = |BC|
> >
b) AD = BC
c) AB k CD
Proof. a) |AD| = |BC| just when △OAD ≡ △OBC (case SSS) just when
>
>
>
\ = AD = BC = BDC
\ = BOC
\ just when b) >
\
AD = BC just when ABD
AOD
2
2
(angles on the circle) just when c) AB k CD.
Lemma 10.12 (Rectangle in a circle). Let ABCD be a quadrilateral whose
vertices are on a circle Cr O. The following are equivalent:
a) ABCD is a rectangle.
b) AC and BD are diameters of the circle (i.e., A, O, D are collinear,
and B, O, C are collinear).
Proof: Exercise.
Theorem 10.13. Let ABCD be a quadrilateral. The following are equivalent:
a) The quadrilateral ABCD is cyclic.
\ = ACD.
\ (The angle formed by a diagonal with a side is equal
b) ABD
with that formed by the other diagonal with the opposite side).
\ + ADC
\ = 180◦ (The sum of two opposite angles is 180◦ ).
c) ABC
b
B
b
C
A
b
b
D
Proof.
a) We assume a) and prove b) and c). On the circle C containing the
vertices A, B, C, D, we have
>
AD
\
\
ABD = ACD =
,
2
> >
◦
\ + ADC
\ = ADC + ABC = 360 = 180◦ .
ABC
2
2
EUCLIDEAN GEOMETRY
37
>
Here ADC denotes the arc bounded by A and C and containing
>
the point D, while ABC denotes the arc bounded by A and C and
containing the point B.
b) Assume b). Prove a). Let C be the circle containing the points A,
B, C. This is the circle whose centre is the circumcenter O of the
triangle ABC (the intersection of the perpendicular bisectors), and
whose radius is |OA|. We would like to prove that D is also a point
on the circle C . Proof by contradiction: assuming D is not on the
circle C , let C intersect the line CD at the point E, the line BD at
the point F .
b
B
b
A
b
b
E
A
b
b
C
D
b
b
C
B
b
b
F
b
F
b
E
D
We have
>
>
AF
AE
\
\
while ACD =
,
ABD =
2
2
\ = ACD,
\
as angles on the circle. By assumption, we know ABD
> >
so by the equations above, AF = AE.
However, if D is outside the circle C , then F is inside the arc
> >
and so AF < AE. Contradiction.
>
However, if D is inside the circle C , then E is inside the arc AF
> >
so AE < AF . Contradiction.
and
>
AE
and
c) Assume c). Prove a). Similar with the previous part. Let C be the
circle containing the points A, B, C. We would like to prove that D
is also a point on the circle C . Proof by contradiction: assuming D
is not on the circle C , let C intersect the line CD at the point E, the
line AD at the point L. We have
>
ALC
\
,
ABC =
2
38
ANCA MUSTATA
as angle on the circle, and
> >
ABC ± EF L
\
,
ADC =
2
as angle which is either internal, or external to the circle. By as\ + ADC
\ = 180◦ , and so by the equations
sumption, we know ABC
above,
> > >
ALC + ABC ± EF L = 360◦ .
However,
> >
ALC + ABC = 360◦
>
as they span the entire circle, so LE = 0, meaning that L = E. But
this would mean that the lines CD and AD intersect the circle at the
same point E = L. As the intersection of AD and CD is D, we must
have D = E = L.
Example 10.14. Let H be the orthocentre of △ABC, and let D′ denote
the symmetric of H through BC. Then D′ is a point on the circumcentre
of △ABC.
b
b
b
E
b
B
A
b
b
b
O
H
D
b
C
D′
Proof. BC is the perpendicular bisector of HD′ =⇒ △CDH ≡ △CDD′ .
′ C = DHC
\
\ = 90◦ − HCD
\ = ABC
\ so the quadrilateral ABD′ C is
Then AD
cyclic.
(We used HD ⊥ BC and CH ⊥ AB. )
11. Similar triangles
Triangles △ABC and △A′ B ′ C ′ are similar if their respective angles are
equal: Â = Â′ , B̂ = B̂ ′ , Ĉ = Ĉ ′ . we write
△ABC ∼ △A′ B ′ C ′
EUCLIDEAN GEOMETRY
39
Theorem 11.1. If △ABC ∼ △A′ B ′ C ′ then
|AB|
|AC|
|BC|
= ′ ′ = ′ ′ .
′
′
|A B |
|A C |
|B C |
As an application, sin α and cos α as defined in trigonometry are independent of the choice of the triangle in which they are computed.
The most common situations when two similar triangles arise are the
following:
Theorem 11.2 (Parallel lines III). BC k B ′ C ′ just when △ABC ∼ △AB ′ C ′ .
b
B′
B
A
b
b
b
C′
b
C
Theorem 11.3 (Anti-parallel lines). If B, C, C ′ , B ′ are on the same circle
and BC ′ intersects CB ′ at A then △ABC ∼ △AB ′ C ′ ; BC and B ′ C ′ are
anti-parallel.
b
b
B′
b
B
A
B
b
b
C
b
A
B
b
b
b
A
b
b
b
C
B′
C
B′
b
′
b
C′
b
C′
Please read the Notes on Areas and solve Ex. Set. 3 for more practice
with similar triangles.
In particular, we can solve the following question: Given a circle Cr O and
a point P , how can we describe how far the point is from the circle?
The power of a point P with respect to a circle is the product of the
segments made by the point P on any chord passing through P .
Question (9) from Ex. Set. 3 can be reformulated as follows:
C
40
ANCA MUSTATA
Theorem 11.4. The power of a point P with respect to a circle Cr O does
not depend on the chord on which it is calculated:
|P A| · |P B| = |P A′ | · |P B ′ | = ±(|P O|2 − r2 ),
+ if P is outside the circle and − if P is inside the circle.
b
′
B
b
B
A
b
b
A
b
A
P
b
b
O
b
P
b
A′
b
b
O
b
B′
B′
Proof. Due to the equal angles on the circle, △P AA′ ∼ △P B ′ B so that
|P A|
|P A′ |
=
|P B ′ |
|P B|
and thus |P A||P B| = |P A′ ||P B ′ | = ±(|OP | − r)(|OP | + r)..
EUCLIDEAN GEOMETRY
41
12. The nine-point circle
E′
A
b
b
B′
C′
b
b
M′
E
b
b
P
N
b
b
O
b
G
O′
b
b
F′
F
b
H
b
b
N′
b
b
P′
D
b
b
b
b
B
M
b
D
b
′
A′
C
42
ANCA MUSTATA
Can you guess the significance of each point? If you were to connect all
labeled points, could you find
• at least 19 segments whose midpoints are labeled in?
• at least 9 perpendicular bisectors?
• at least 15 parallelograms which are not rectangles?
• at least 9 isosceles trapezoids?
• at least 6 diameters and 9 rectangles?
• at least 9 pairs of similar triangles sharing H as common vertex?
• at least 12 pairs of similar triangles sharing A as common vertex?
• at least 8 triangles having G as centroid?
13. Menelaus and Ceva theorems
Theorem 13.1 (Menelaus). Let M, P be points inside the segments [AB],
and [CA], and let N be a point on the line BC, outside of the segment [BC].
M, N, P are collinear if and only if
|AM | |BN | |CP |
=1
|M B| |N C| |P A|
A
b
M
b
P
b
B
b
b
N
b
C
Proof. If the points are collinear, then to prove relation: Let CD k AB.
b
A
M
b
b
P
b
B
b
b
D
C
b
N
|
|M B|
|CP |
|CD|
Then multiply |BN
|N C| = |CD| with |P A| = |AM | .
If we know the relation, assume the points are not collinear, let M ′ be
the intersection of lines N P and AB. It has to lie inside the segment [AB]
just like M . Then by the previous argument,
|AM ′ | |BN | |CP |
=1
|M ′ B| |N C| |P A|
EUCLIDEAN GEOMETRY
43
while by assumption
|AM | |BN | |CP |
=1
|M B| |N C| |P A|
′
|AM |
|AM |
hence |M
′ B| = |M B| so
|AM ′ | thus M is M ′ .
|AM ′ |
|AB|
=
|AM ′ |
|AM ′ |+|M ′ B|
=
|AM |
|AM |+|M B|
=
|AM ′ |
|AB|
so |AM | =
Theorem 13.2 (Ceva). Let M, N, P be points inside the segments [AB],
[BC] and [CA]. The lines AN, BP and CM are concurrent if and only if
|AM | |BN | |CP |
= 1.
|M B| |N C| |P A|
Proof.
a) We assume the lines AN, BP and CM are concurrent. We apply
Menelaus’ Theorem twice: once for triangle ABN crossed by line
CM and then for triangle ACN crossed by line BP . Multiplying the
two ensuing relations yields the formula above.
b) We assume the formula above. We let Q be the intersection point of
lines BP and CM . Let N ′ denote the intersection of line AQ with
BC. From Part I, we get:
|AM | |BN ′ | |CP |
= 1.
|M B| |N ′ C| |P A|
From assumption, we have
|AM | |BN | |CP |
= 1.
|M B| |N C| |P A|
Together, these yield
|BN ′ |
|BN |
=
.
′
|N C|
|N C|
so P = P ′ since both P and P ′ are inside the segment [BC].
References
[1] Michèle Audin, Geometry, Springer-Verlag, Heidelberg, 2003.
[2] H.S.M. Coxeter, Introduction to Geometry, 2ed., Wiley, 1969.
[3] Euclid, The Elements, on-line editions:
http://farside.ph.utexas.edu/euclid/Elements.pdf
http://www.gutenberg.org/files/21076/21076-pdf.pdf
[4] Geometry Package GeoGebra
http://www.geogebra.org/cms/
[5] Robin Hartshorne, Geometry: Euclid and Beyond, Springer-Verlag, Heidelberg, 2000.
[6] Silvio Levy, preface to Flavours of Geometry, MSRI, Berkeley, 1997.
Index
Euler’s line, 32
exterior angle, 5
acute angle, 4
acute-angled triangle, 5
altitude, 25
theorem, 28
angle, 4
acute, 4
alternate, 10
corresponding, 10
interior consecutive, 10
obtuse, 4
angle bisector, 21
theorem, 22
angle-side-angle, 14
anti-parallel, 39
arc, 35
ASA, 14
axiom, 2
geometric locus, 22
half-line, 3
half-planes, 7
hexagon, 5
hypotenuse, 5
hypothesis, 2
in between, 6
incentre, 23
theorem, 23
incircle, 23
interior, 6
interior angle, 5
isosceles, 5
isosceles triangle
theorem, 14
centre, 4
centroid, 30
theorem, 30
Ceva theorem, 43
circle, 4
nine-point, 41
circumcentre, 24
circumcircle, 24
circumference, 4
cirumcentre
theorem, 24
collinear, 3
complement, 4
conclusion, 2
concurrent, 3
concyclic, 5
congruent, 6, 13
convex, 7
corollary, 2
cosine, 27
cyclic, 35
leg, 4
lemma, 2
line
parallel, 5
locus, 22
mathematical theory, 2
measure, 6
median, 29
theorem, 31
Menelaus theorem, 42
midline, 29
nine-point circle, 41
obtuse angle, 4
obtuse-angled triangle, 5
orthocenter, 25
orthocentre
theorem, 25
parallel, 9
parallel line, 5
parallelogram, 16
pentagon, 5
perpendicular, 4
perpendicular bisector, 23
theorem, 24
definition, 2
diameter, 5
distance
from point to line, 21
equilateral, 5
Euler, 32
44
EUCLIDEAN GEOMETRY
plane, 4
point, 3
polygon, 5
regular, 11
power, 39
projection, 33
proof, 2
proposition, 2
Pythagorean theorem, 19
quadrilateral, 5
radius, 5
rays, 3
rectangle, 18
theorem, 18
rhombus, 18
rhombus theorem, 18
right angle, 4
right-angled triangle, 5
rigid movement, 6
SAS, 13
scalene, 5
secant, 33
segment, 3
side, 5
side-angle-side, 13
side-side-side, 15
similar triangles, 38
sine, 27
smaller, 6
SSS, 15
straight line, 3
sum, 4
sum of angles in a triangle, 11
supplement, 4
surface, 4
tangent, 23, 33
circles, 34
tessellation, 11
theorem, 2
altitude, 28
angle bisector, 22
angle-side-angle, 14
centroid, 30
Ceva, 43
circumcentre, 24
incentre, 23
isosceles triangle, 14
median, 31
Menelaus, 42
orthocentre, 25
parallel lines I, 9
parallel lines II, 10
parallel lines III, 39
parallelogram, 16
perpendicular bisector, 24
Pythagorean, 19
rectangle, 18
rhombus, 18
side-angle-side, 13
side-side-side, 15
sum of angles in a triangle, 11
tiling, 11
triangle, 5
vertex, 4
45