Download Chapter 3 Foundations of Geometry 2

Chapter 3
Foundations of Geometry 2
3.1
Triangles, Congruence Relations, SAS Hypothesis
Definition 3.1 A triangle is the union of three segments ( called its side), whose end
points (called its vertices) are taken, in pairs, from a set of three noncollinear points. Thus,
if the vertices of a triangle are A, B, and C, then its sides are AB, BC, AC, and the triangle
is then the set defined by AB ∪ BC ∪ AC, denoted by ∆ABC. The angles of ∆ABC are
∠A ≡ ∠BAC, ∠B ≡ ∠ABC, and ∠C ≡ ∠ACB.
Equality and Congruence
Equal(=): identically the same as. For examples,
1) Two points are equal, as in A = B, we mean they coincide.
2) AB = CD only if the set of points AB is the exact same set of points denoted by CD.
Congruence(∼
=): is an equivalence relation. (to be defined later).
Congruence for Segments and angles
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AB ∼
iff AB = XY
= XY
∠ABC ∼
= ∠XY Z iff m∠ABC = m∠XY Z
Congruence for triangles
Notation: correspondence between two triangles Given ∆ABC and ∆XY Z, write
ABC ↔ XY Z
to mean the correspondence between vertices, sides and angles in the order written.
Note: There are possible six ways for one triangle to correspond to another.
ABC ↔ XY Z
ABC ↔ XZY
ABC ↔ Y XZ
ABC ↔ Y ZX
ABC ↔ ZXY
ABC ↔ ZY X
Definition 3.2 (Congruence for triangles) If, under some correspondence between the
vertices of two triangles, corresponding sides and corresponding angles are congruent, the
triangles are said to be congruent. Thus we write ∆ABC ∼
= ∆XY Z whenever
AB ∼
= XY , BC ∼
= Y Z, AC ∼
= XZ,
∠A ∼
= ∠X, ∠B ∼
= ∠Y, ∠C ∼
= ∠Z
Notation: CPCF: Corresponding parts of congruent figures (are congruent).
Properties of Congruence
1. reflexive law: ∆ABC ∼
= ∆ABC
2. Symmetry Law: If ∆ABC ∼
= ∆XY Z, then ∆XY Z ∼
= ∆ABC 3. Transitive Law: If
∼
∆ABC = ∆XY Z, and ∆XY Z ∼
= ∆U V W , then ∆ABC ∼
= ∆U V W .
Remark: Euclid did not use the word congruence. Euclid attributed to congruent triangles
the property that on triangle could be placed precisely on top of another.
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3.2
Chapter 3. Foundations of Geometry 2
27
Taxicab Geometry: Geometry without SAS congruence
Skip
3.3
SAS, ASA, SSS Congruence and Perpendicular Bisectors
Question: Can we require fewer than six sets of congruent pairs to determine two triangles
are congruent? The SAS Hypothesis
Under the correspondence ABC ↔ XY Z, let two sides and the included angles of ∆ABC
be congruent, respectively, to the corresponding two sides and the included angle of ∆XY Z.
Note: This can not be established within the current set of axioms. (See Section 3.2.
Axiom 3.3 (SAS Postulate) If the SAS Hypothesis holds for two triangles under some
correspondence between their vertices, then the triangles are congruent.
Theorem 3.4 (ASA Theorem) If, under some correspondence, two angles and the included side of one triangle are congruent to the corresponding angles and included side of
another, the triangles are congruent under that correspondence.
Proof: Outline of the proof: (Use SAS postulate).
Given ∠A ∼
= ∠X, AB ∼
= XY , and ∠B ∼
= ∠Y . If AC ∼
= XZ, then the two triangles are
congruent by SAS. But if not, then either AC > XZ or AC < XZ.
1) Assume AC > XZ, then we can find D on AC such that A − D − C and AD ∼
= XZ by
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Chapter 3. Foundations of Geometry 2
the Segment Construction Theorem. This will leads to a contradiction.
2) One can argue similarly for the case AC < XZ.
Example 3.5 Given: ∠EBA ∼
= ∠CBD, AB ∼
= BC, ∠A ∼
= ∠C.
∼
Prove: EB = DB.
Conclusions
(1) ∆ABE ∼
= ∆CBD
(2) ∴ EB ∼
= DB
Example 3.6
Justifications
ASA
CPCF
28
2
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Given: M is the midpoint of CD and EF .
Prove: ∠C = ∠D.
Example 3.7 Prove that a line that bisects an angle also bisects any segment perpendicular
to it that joins two points on the sides of that angle.
Isosceles Triangle Theorem
Definition 3.8 Isosceles triangle: a triangle having two sides congruent.
Legs, base, base angles, vertex, vertex angle.
Lemma 3.9 In ∆ABC, if AC ∼
= BC, then ∠A ∼
= ∠B.
Proof: By the SAS postulate: ∆CAB ∼
= ∆CBA, therefore, corresponding angles ∠CAB
and ∠CBA are congruent, that is ∠A ∼
= ∠B.
(You could also construct an angle bisector from the vertex).
2
Theorem 3.10 A triangle is isosceles iff the base angles are congruent.
Proof: Only need to show the converse. By the ASA theorem, ∆CAB ∼
= ∠CBA. Therefore
AC ∼
BC.
=
2
Example 3.11 Solve the following problem in proof writing.
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Given: GJ = KM = 2, JK = HJ = HK = 10, with betweenness relations as evident from
the figure. Prove: HG = HM .
Symmetry
←−→
Lemma 3.12 If M is the midpoint of segment AB and line P M is perpendicular to AB,
then P A = P B.
Lemma 3.13 If P A = P B and M is the midpoint of segment AB, then line P M is perpendicular to segment AB.
Lemma 3.14 If P A = P B and M is the midpoint of segment AB, then line P M bisects
∠AP B.
Perpendicular Bisectors, Locus
Definition 3.15 The Perpendicular bisector of a segment AB to be the line that both
bisects AB and is perpendicular to it.
Locus: The locus of a point is the path or set of points that is determined by that point
when it satisfies certain given properties.
Theorem 3.16 (Perpendicular Bisector Theorem) The set of all points (The locus of
a point which is ) equidistant from two distinct points A and B is the perpendicular bisector
of the segment AB.
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Remark: To prove above theorem, actually one needs to prove two sets are equal. Namely,
the Locus (set of points, in this theorem, it is the perpendicular bisector) is equal to the set
of points which are equidistant from the two distinct points A and B.
Proof: (⇒) Let ` be the perpendicular bisector of AB, and assume P ∈ `. By Lemma 3.12,
P A = P B.
(⇐): Assume P is equidistant from A and B. Let M be the midpoint of AB, then by Lemma
←−→
←−→
3.13, P M ⊥AB. (Next argue P M = `). But `⊥AB, there can be only one perpendicular to
←−→
AB at M . Hence P M = `, and P ∈ `.
2
Theorem 3.17 If, under some correspondence between their vertices, two triangles have the
three sides of one congruent to the corresponding three sides of the other, then the triangles
are congruent under that correspondence.
Proof: class exercise. Note, there are three different cases, but can be reduced to two cases.
2
Example 3.18 Name all congruent pairs of distinct segments and angles in the following
figure.
Example 3.19 In the following figure, we have a four-sided figure with the congruent sides
and right angles as marked.
(a) Prove that AD ∼
= CD.
(b) If, in addition, BC ∼
= CD, prove ∠D is a right angle.
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Theorem 3.20 (Existence of perpendicular from an external point) Let the line `
and point A not on ` be given. Then there exists a unique line m perpendicular to ` passing
through A.
Proof: (Existence) By construction. Locate B and C on `. let ∠DBC ∼
= ∠ABC, BD ∼
= BA.
It follows that B and C are both equidistant from A and D. (Why) Hence AD⊥BC by the
perpendicular bisector theorem.
(Uniqueness): Class discussion.
2
3.4
Exterior Angle Inequality
←→
Definition 3.21 Let ∆ABC be given, and suppose D is a point on BC such that the betweenness relation B − C − D holds. Then ∠ACD is called an exterior angle of the given
triangle. The angles at A and B of ∆ABC are called opposite interior angles of ∠ACD.
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Remark:
1) In Euclidean geometry: The measure of an exterior angle of any triangle equals the sum
of the measures of the other two opposite interior angles.
2) in absolute geometry: the above relationship is not valid in general.
Definition 3.22 Absolute geometry consists of part of Euclidean geometry that includes
all the axioms except all references to parallel lines, or results therefrom. Absolute geometry
provides foundations not only for Euclidean geometry, but also for non-Euclidean geometry.
Exterior angles: example provided by Henri Poincarè
Poincaré Model for Absolute Geometry
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1. C is a circle with center O;
2. All points inside C are “points” of this geometry. A point on or outside the circle is
not a “point”;
3. “line”: either an straight line through O, cut off by C, or the arc of a circle that makes
right angle with C and the arc lies inside C;
4. Betweeness: If on an arc of a circle point Q is between points P and R, then we define
P − Q − R. And the definitions for segments, rays, and angles follow this betweenness
definition.
5. for example, the “segment” AB shown in the following figure.
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6. Angle measure: the angle between two curves defined as the angle measure of the angle
formed by the tangents to the two curves at the point of intersection.
7. The geometry inside C satisfies all the axioms for absolute geometry in plane. For
example: two points determine a unique line.
8. Consider one of the triangles: (see the following figure)
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Chapter 3. Foundations of Geometry 2
The angle sum of a triangle in this geometry is always less than 180.
9. SAS Postulate holds for Poincarè model.
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External angles in Spherical Geometry
1. Take the unit sphere S;
2. “points”: all the points on the surface of the sphere;
3. “lines”: all great circles of that sphere; “equator”: the great circle on S that lies in a
horizontal plane;
“meridian”: the great circles passing through the north and south poles;
4. distance: ordinary (Euclidean) arc length;
5. angle measure: use the measure of the angle between the tangents to the sides at the
vertex of angle
6. the axioms of absolute geometry work in this geometry, such as two points determine
a unique line.
7. triangle: (a spherical triangle) is simply one made up of arcs of great circles, pairwise
connected at the endpoints.
8. SAS Postulate holds for spherical geometry.
9. the angle sum of a triangle is always greater than 180.
Exterior angle of a triangle in absolute geometry
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Theorem 3.23 (exterior angle Inequality) An exterior angle of a triangle has angle
measure greater than that of either opposite interior angle.
Remark: This theorem is true for absolute geometry. That means it is true without the
concept of parallelism. It’s a weak form of its Euclidean counterpart.
Proof: See above figure.
2
Applications
Corrolary 3.24 1) The sum of the measures of any two angles of a triangle is less than
180.
2) A triangle can have at most one right or obtuse angle.
3) The base angles of an isosceles triangle are acute.
Proof: Proof of the first statement. Use exterior angle inequality.
2
Example 3.25 Consider the triangle shown below, with certain angle measures indicated.
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(a) Use the exterior Angle Inequality to show that ∠C has measure less than 131.
(b) In this example, is the angle sum of ∆ABC equal to, or less than 180? (Do not use
Euclidean geometry. )
Example 3.26 If ∠ECD is an exterior angle of ∆EAC and A − B − C − D holds, use the
Exterior Angle Inequality to find upper and lower bounds for
(a) m∠EBC = x
(b) m∠BEC = y
Theorem 3.27 (Saccheri-Legendre Theorem) The angle sum of angle triangle can not
exceed 180.
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Proof: Outline of the proof:
1) Let ∆ABC be any given triangle. Locate the midpoint M of AC then extend BM to
E such that BM = M E. Repeat this construction in ∆BEC. Continue this process at
infinitum.
Lemma 3.28 the angle sums of all the new triangles constructed in the process remain constant.
2) Assume the angle sum of ∆ABC greater than 180, i.e., there is a constant t > 0, such
that
m∠A + m∠ABC + m∠BCA = 180 + t
3) The measures of angles at E, F, G, · · · are decreasing. Hence eventually have measure
< t.
Observe that
θ1 + θ2 + θ3 + · · · + θn < m∠ABC
thus, there exists a n large enough, such that
θn < t
4) Assume when θn < t, the corresponding triangle is ∆BCW . By 2) The angle sum of
∆BCW = 180 + t, so it follows that
180 + t = m∠W BC + m∠BCW + m∠W < m∠W BC + m∠BCW + t
i.e.,
180 < m∠W BC + m∠BCW,
which is a contradiction.
2
Remark: We haven’t yet proven that the angle sum of a triangle is 180. That needs the
Parallel Postulate.
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3.5
Chapter 3. Foundations of Geometry 2
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The Inequality Theorems
Theorem 3.29 (Scalene Inequality) If one side of a triangle has greater length than
another side, then the angle opposite the longer side has the greater angle measure, and,
conversely, the side opposite an angle having the greater measure is the longer side.
Proof: Outline:
1) In ∆ABC it is given that AC > AB. Locate D on AC so that AD = AB, and joint
points B and D.
m∠ABC > m∠1 = m∠2 > m∠C
2)Conversely, Given m∠B > m∠C. Assume AC < AB, then by (1), m∠B < m∠C, which
is a contraction.
2
Corrolary 3.30 (1) If a triangle has an obtuse or right angle, then the side opposite that
angle has the greatest measure.
(2) The hypotenuse of a right triangle has measure greater than that of either leg.
Theorem 3.31 If A, B and C are any three distinct points, then AB + BC ≥ AC, with
equality only when the points are collinear, and A − B − C.
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Proof: Outline:
1) When A, B, and C are collinear. Without loss of generality, assume A is to the left of C.
Either A − B − C, B − A − C or A − C − B. Discuss each case.
2) Consider the noncollinear case.
Extend CB to D such that BD = BA. In ∆DAC, DC = AB + BC > AC by the Scalene
inequality.
2
Corrolary 3.32 (Median Inequality) Suppose that AM is the median to side BC of
∆ABC. Then
1
AM < (AB + AC)
2
Proof: Exercise.
2
Example 3.33 Find which of the angle measures x, y, z, r, and s indicated in the following
figure is the leaset. Prove your answer.
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Theorem 3.34 (SAS Inequality Theorem) If in ∆ABC and ∆XY Z we have AB =
XY, AC = XZ, but m∠A > m∠X, then BC > Y Z, and conversely if BC > Y Z, then
m∠A > m∠X.
This theorem is also called “Hinge” or “Alligator” Theorem. Proof:
−−→
−→ −−→ −→
Outline: 1) Construct ray AD such that AB − AD − AC and ∠BAD ∼
= ∠X, with AD =
XZ = AC.
2) Construct the angle bisector of ∠DAC; 3) Converse argument is by contradiction.
2
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Example 3.35 In the following figure, a circle with diameter QR and center O is shown.
←→
If P varies on the circle on either side of QR, and if θ = m∠P OQ, define the function
f (θ) = P Q,
0 < θ < 180
Explain why f (θ) is an increasing function. (that is , if θ1 < θ2 , then f (θ1 ) < f (θ2 ).)
Observe points P and P 0 in the figure. We need to prove that P Q, which is f (θ1 ), is less
than P 0 Q, which is f (θ2 ).
Solution: Use SAS inequality theorem.
3.6
Additional Congruence Criteria
Theorem 3.36 (AAS congruence criterion) If under some correspondence between their
vertices, two angles and a side opposite in one triangle are congruent to the corresponding
two angles and side of a second triangle, then the triangles are congruent.
Proof: Outline of proof:
show that the third angle of one triangle must be congruent to the corresponding angle in
the other triangle. This is done by assuming the contrary and the Exterior Angle Inequality.
2 Noncongruent triangles satisfying SSA Hypothesis
1) When conditions of SSA are given, it may have one, two or none solutions for the triangle.
Why?
2) for the case when two triangles satisfying the given SSA condition but not congruent, we
obviously have the following lemma.
Theorem 3.37 (SSA Theorem) If, under some correspondence between their vertices,
two triangles have two pairs of corresponding sides and a pair of corresponding angles congruent, and if the triangles are not congruent under this correspondence, then the remaining
pair of angles not included by the congruent sides are supplementary.
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Corrolary 3.38 If under some correspondence of their vertices, two acute angled triangles
have tow sides and an angle opposite one of them congruent, respectively, to the corresponding
two sides and angle of the other, the triangles are congruent.
The Right Triangle Congruence Criteria
Corrolary 3.39 (HL Theorem) If two right triangles have the hypotenuse and leg of one
congruent, respectively, to the hypotenuse and leg of the other, the right triangles are congruent.
Proof: class discussion.
2
Corrolary 3.40 (HA Theorem) If two right triangles have the hypotenuse and acute angle of one congruent, respectively, to the hypotenuse and acute angle of the other, the triangles
are congruent.
Corrolary 3.41 (LA Theorem) If under some correspondence between their vertices, two
right triangles have a leg and acute angle of one congruent, respectively, to the corresponding
leg and acute angle of the other, the triangles are congruent.
Corrolary 3.42 (SsA congruence Criterion) Suppose that in ∆ABC and ∆XY Z, AB ∼
=
∼
∼
∼
XY , BC = Y Z, ∠A = ∠X, and BC ≥ BA. Then ∆ABC = ∆XY Z.
Proof: class discussion.
Example 3.43 Given: P Q⊥P R, QS⊥SR, and P Q ∼
= QS. Prove: P R ∼
= RS.
2
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Definition 3.44 The distance between two geometric objects (set of points) is the distance
between the the closest points in the two sets.
Definition 3.45 The distance from any point P to a line ` not passing through P is the
distance from P to the foot of the perpendicular Q from P to line `. A point is equidistant
from two lines iff the distances from the point to the two lines are equal.
Theorem 3.46 The distance form a point P to any point Q in line ` is least when P Q⊥`.
Proof: class discussion.
2
Example 3.47 Prove that if P A = P B and M is the midpoint of AB, then M is equidistant
−→
−→
from rays P Q and P R.
3.7
Quadrilaterals
Notation:
1: 3: quadrilateral
2. 2: Squares or rectangles
3. : Parallelogram
Definition 3.48 If A, B, C and D are any four points lying in a plane such that no three of
them are collinear, and if the points are so situated that no pair of open segments determined
by each pair of points taken in the order A, B, C and D (AB, BC, etc.) have points in
common, then the set
3ABCD ≡ AB ∪ BC ∪ CD ∪ DA
is a quadrilateral, with vertices A, B, C, D, sides AB, BC, CD, DA, diagonals AC, BD,
and angles ∠DAB, ∠ABC, ∠BCD, ∠CDA.
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adjacent(or consecutive) sides or angles
opposite sides or angles
Convex quadrilaterals
Definition 3.49 A quadrilateral with its diagonals intersecting at a point that lies between
opposite vertices.
Properties of a convex quadrilateral
• The diagonals of a convex quadrilateral intersect at an interior point on each diagonal;
• if 3ABCD is a convex quadrilateral, then D lies in the interior of ∠ABC, and similarly
for the other vertices;
• If A, B, C, and D are consecutive vertices of a convex quadrilateral, then m∠BAD =
m∠BAC + m∠CAD.
Congruence criteria for convex quadrilaterals
Definition 3.50 Two quadrilaterals 3ABCD and 3XY ZW are congruent under the correspondence ABCD ↔ XY ZW iff all pairs of corresponding sides and angles under the
correspondence are congruent (i.e., CPCF). Such congruence will be denoted by
3ABCD ∼
= 3XY ZW
.
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Theorem 3.51 (SASAS Congruence) Suppose that two convex quadrilaterals 3ABCD
and 3XY ZW satisfy the SASAS Hypothesis under the correspondence ABCD ↔ XY ZW .
That is, three consecutive sides and the the two angles included by those sides of 3ABCD
are congruent, respectively, to the corresponding three consecutive sides and two included
angles of 3XY ZW . Then 3ABCD ∼
= 3XY ZW .
Proof: We must prove that the remaining corresponding sides and angles of the two quadrilaterals are congruent.
2 Other congruence theorems for convex quadrilaterals are
ASASA Theorem
SASAA Theorem
SASSS Theorem
Question: Is ASSSS a valid congruence criterion for convex quadrilaterals?
Saccheri, Lambert quadrilaterals
Definition 3.52 A rectangle is a convex quadrilateral having four right angles.
Definition 3.53 Let AB be any line segment, and erect two perpendiculars at the endpoints
A and B. Mark off points C and D on these perpendiculars so that C and D lie on the dame
side of line AB, and BC = AD. The resulting quadrilateral is a Saccheri Quadrilateral.
Side AB is called the base, BC and AD the legs, and side CD the summit. The angles
at C and D are called the summit angles.
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Saccheri Quadrilateral in non-Euclidean Geometry
Remark:
1) A Saccheri Quadrilateral in the Poincarè Model has acute summit angles;
2) A Saccheri Quadrilateral on the unit sphere has obtuse summit angles;
←
→
←→
Lemma 3.54 lines BC and Ad in the Saccheri quadrilateral can not meet.
Proof: Because the uniqueness of perpendiculars from an external point in absolute geometry.
2
Lemma 3.55 A Saccheri Quadrilateral is convex.
Why?
Theorem 3.56 The summit angles of a Saccheri Quadrilateral are congruent.
Proof: See the following picture. 3DABC ∼
= 3CBAD under the correspondence DABC ↔
CBAD by SASAS Theorem.
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2
Corrolary 3.57 1) The diagonals of a Saccheri Quadrilateral are congruent.
2) The line joining the midpoints of the base and summit of a Saccheri Quadrilateral is the
perpendicular bisector of both the base and summit.
3) If each of the summit angles of a Saccheri Quadrilateral is a right angle, the quadrilateral
is a rectangle, and the summit is congruent to the base.
Proof: class discussion. 1) and 3) are trivial.
2
Definition 3.58 (Lambert quadrilateral) A quadrilateral in absolute geometry having
three right angles is called a lambert Quadrilateral.
Remark: its existence is guaranteed by the above Corollary.
Three possible hypothesis
1) Summit angles of a Saccheri Quadrilateral are obtuse;
2) Summit angles of a Saccheri Quadrilateral are right angles;
3) Summit angles of a Saccheri Quadrilateral are acute;
Theorem 3.59 The Hypothesis of the Obtuse Angle is not valid in absolute geometry.
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Outline of the proof: (By construction):
Locate the midpoints M and N of sides AB and AC of any triangle ABC, and draw line
←−→
` = M N . Then drop perpendiculars BB 0 and CC 0 from B and C to line `.
(1) Show 3BCC 0 B 0 (called the Saccheri Quadrilateral associated with ∆ABC) is a
Saccheri Quadrilateral. (BB 0 = CC 0 and congruent summit angles at B and C. )
(2) The angle sum of ∆ABC has twice the value of the measure of x of each summit angle.
thus
2x ≤ 180
(3) For any Saccheri Quadrilateral there is an associated triangle. How?
Remark: 1) The Hypothesis of the Acute Angle for Saccheri is also false. But it is impossible
to prove this with only the axioms of absolute geometry.
2) The length of the base B 0 C 0 equals twice the length of M N .
3) The summit of a Saccheri Quadrilateral has length greater than or equal to that of the
base.
4) The line segment joining the midpoints of two sides of a triangle has length less than or
equal to one-half of the third side.
3.8
Circles
Definition 3.60 A circle is the set of points in a plane that lies at a positive, fixed distance
r from some fixed point O. The number r is called the radius, and the fixed point O is
called the center of the circle. A point P is said to be interior to the circle, or an interior
point, whenever OP < r; if OP > r, then P is said to be an exterior point.
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Other terminologies
See the following picture:
Elementary properties of a circle
• The center of a circle is the midpoint of any diameter.
• The perpendicular bisector of any chord of a circle passes through the center.
• A line passing through the center of a circle and perpendicular to a chord bisects the
chord.
• A line passing through the center of a circle and perpendicular to a chord bisects the
chord.
• Two congruent central angles subtend congruent chords, and conversely.
• Two chords equidistant from the center of a circle have equal lengths, and conversely.
Circular Arc Measure
Definition 3.61 A minor arc is the intersection of the circle with a central angle and its
interior, a semicircle is the intersection of the circle with a closed half-plane whose edge
passes through the center of the circle, and a major arc of a circle is the intersection of
ú of the arc as
the circle and a central angle and its exterior. We define the measure mACB
follows: (see the following figure)
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Minor Arc
ú = m∠AOB
mACB
Chapter 3. Foundations of Geometry 2
53
Semicircle
Major Arc
ú
ú = 360 − m∠AOB
mACB = 180 mACB
ú
ú
Theorem 3.62 (Additivity of Arc Measure) Suppose arcs A1 = AP
B and A2 = BQC
are any two arcs of circle O having just one point B in common and such that their union
ú is also an arc. Then m(A ∪ A ) = mA + mA .
A1 ∪ A2 = ABC
1
2
1
2
Remark: Observe that if we are given two arcs on a circle, one of them has to be a minor
arc.
Outline of the proof:
We distinguish three cases:
ú is a minor arc or a semicircle.
1) when ABC
ú is a major arc and BQC
ú is either a minor arc or a semicircle.
2) When ABC
ú and BQC
ú are major arcs. See the following picture:
3) When both ABC
ú
Example 3.63 Arc SP
T shown in the following picture is a major arc and is the union of
ú
ú
arc SV R (a minor arc) and arc RP
T (a major arc). Using the angle measures shown in
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the figure, determine the measures of each of the three arcs, and verify the additivity in this
case.
Definition 3.64 A line that meets a circle in two distinct points is a secant of that circle.
A line that meets a circle at only one point is called a tangent to the circle, and the point
in common between them is the point of contact, or point of tangency.
Theorem 3.65 (Tangent Theorem) A line is tangent to a circle iff it is perpendicular
to the radius at the point of contact.
Proof: 1) Assume a line ` is tangent to a circle with center C at A, then A is the point of
contact. Need to prove AC⊥`. (How?)
2) Conversely, assume AC⊥`, show A is the only contacting point.
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Corrolary 3.66 If two tangents P A and P B to a circle O from a common external point
−→
P have A and B as the points of contact with the circle, then P A ∼
= P B and P O bisects
∠AP B.
Theorem 3.67 (Secant Theorem) If a line ` passes through an interior point A of a
circle, it is a secant of the circle and intersects that circle in precisely two points.
Outline of the proof:
1. Use Intermediate Value Theorem to prove the line ` intersects the circle at one point.
2. Then use an elementary geometry construction to prove there is another intersection
point. (construct congruent triangles).
3. Show no other intersection points.
Remark:: 1. The Secant Theorem proves, that a line segment joining a point inside a circle
with a point outside must intersect the circle.
Yi Wang
Chapter 3. Foundations of Geometry 2
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2. In general, this is true for any simple closed curves (called Jordan curve) . (Jordan
Closed Curve Theorem).