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Home , Absolute geometry, Congruence (geometry), Foundations of geometry, Right angle

Chapter 3

Foundations of 2

3.1 , Relations, SAS Hypothesis

Definition 3.1 A 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 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

25 Yi Wang Chapter 3. 2 26

AB ∼= XY iff AB = XY ∠ABC ∼= ∠XYZ iff m∠ABC = m∠XYZ Congruence for triangles Notation: correspondence between two triangles Given ∆ABC and ∆XYZ, write

ABC ↔ XYZ

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 ↔ YXZ

ABC ↔ Y ZX ABC ↔ ZXY ABC ↔ ZYX

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 ∼= ∆XYZ whenever

AB ∼= XY, BC ∼= YZ, 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. Law: If ∆ABC ∼= ∆XYZ, then ∆XYZ ∼= ∆ABC 3. Transitive Law: If ∆ABC ∼= ∆XYZ, and ∆XYZ ∼= ∆UVW , then ∆ABC ∼= ∆UVW . Remark: did not use the word congruence. Euclid attributed to congruent triangles the property that on triangle could be placed precisely on top of another. Yi Wang Chapter 3. Foundations of Geometry 2 27

3.2 : Geometry without SAS con- gruence

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3.3 SAS, ASA, SSS Congruence and Bi- sectors

Question: Can we require fewer than six sets of congruent pairs to determine two triangles are congruent? The SAS Hypothesis Under the correspondence ABC ↔ XYZ, let two sides and the included angles of ∆ABC be congruent, respectively, to the corresponding two sides and the included of ∆XYZ. Note: This can not be established within the current set of . (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 ) If, under some correspondence, two angles and the in- cluded 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 Yi Wang Chapter 3. Foundations of Geometry 2 28

the Segment Construction Theorem. This will leads to a contradiction. 2) One can argue similarly for the case AC < XZ. 2

Example 3.5 Given: ∠EBA ∼= ∠CBD, AB ∼= BC, ∠A ∼= ∠C. Prove: EB ∼= DB.

Conclusions Justifications (1) ∆ABE ∼= ∆CBD ASA (2) ∴ EB ∼= DB CPCF

Example 3.6 Yi Wang Chapter 3. Foundations of Geometry 2 29

Given: M is the midpoint of CD and EF . Prove: ∠C = ∠D.

Example 3.7 Prove that a 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 : a triangle having two sides congruent. Legs, base, base angles, , 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. Yi Wang Chapter 3. Foundations of Geometry 2 30

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 PM is perpendicular to AB, then PA = PB.

Lemma 3.13 If PA = PB and M is the midpoint of segment AB, then line PM is per- pendicular to segment AB.

Lemma 3.14 If PA = PB and M is the midpoint of segment AB, then line PM 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 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. Yi Wang Chapter 3. Foundations of Geometry 2 31

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, PA = PB. (⇐): Assume P is equidistant from A and B. Let M be the midpoint of AB, then by Lemma ←−→ ←−→ 3.13, PM⊥AB. (Next argue PM = `). But `⊥AB, there can be only one perpendicular to ←−→ AB at M. Hence PM = `, 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 , BC ∼= CD, prove ∠D is a . Yi Wang Chapter 3. Foundations of Geometry 2 32

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 be- tweenness 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. Yi Wang Chapter 3. Foundations of Geometry 2 33

Remark: 1) In : The measure of an exterior angle of any triangle equals the sum of the measures of the other two opposite interior angles. 2) in : 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 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 Poincar`e Poincar´eModel for Absolute Geometry Yi Wang Chapter 3. Foundations of Geometry 2 34

1. C is a 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. Yi Wang Chapter 3. Foundations of Geometry 2 35

6. Angle measure: the angle between two 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 . For example: two points determine a unique line.

8. Consider one of the triangles: (see the following figure) Yi Wang Chapter 3. Foundations of Geometry 2 36

The angle sum of a triangle in this geometry is always less than 180.

9. SAS Postulate holds for Poincar`emodel. Yi Wang Chapter 3. Foundations of Geometry 2 37

External angles in

1. Take the unit S; 2. “points”: all the points on the surface of the sphere; 3. “lines”: all great 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. : ordinary (Euclidean) arc ; 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 Yi Wang Chapter 3. Foundations of Geometry 2 38

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. Yi Wang Chapter 3. Foundations of Geometry 2 39

(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. Yi Wang Chapter 3. Foundations of Geometry 2 40

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 = ME. 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 con- stant.

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∠WBC + m∠BCW + m∠W < m∠WBC + m∠BCW + t i.e., 180 < m∠WBC + 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 . Yi Wang Chapter 3. Foundations of Geometry 2 41

3.5 The Inequality

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 of a 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. Yi Wang Chapter 3. Foundations of Geometry 2 42

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. Yi Wang Chapter 3. Foundations of Geometry 2 43

Theorem 3.34 (SAS Inequality Theorem) If in ∆ABC and ∆XYZ 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 Yi Wang Chapter 3. Foundations of Geometry 2 44

Example 3.35 In the following figure, a circle with QR and center O is shown. ←→ If P varies on the circle on either side of QR, and if θ = m∠POQ, 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).) 0 Observe points P and P in the figure. We need to prove that PQ, which is f(θ1), is less 0 than P 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 con- gruent, and if the triangles are not congruent under this correspondence, then the remaining pair of angles not included by the congruent sides are supplementary. Yi Wang Chapter 3. Foundations of Geometry 2 45

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 con- gruent.

Proof: class discussion. 2

Corrolary 3.40 (HA Theorem) If two right triangles have the hypotenuse and acute an- gle 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 ∆XYZ, AB ∼= XY , BC ∼= YZ, ∠A ∼= ∠X, and BC ≥ BA. Then ∆ABC ∼= ∆XYZ.

Proof: class discussion. 2

Example 3.43 Given: PQ⊥PR, QS⊥SR, and PQ ∼= QS. Prove: PR ∼= RS. Yi Wang Chapter 3. Foundations of Geometry 2 46

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 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 PQ⊥`.

Proof: class discussion. 2

Example 3.47 Prove that if PA = PB and M is the midpoint of AB, then M is equidistant −→ −→ from rays PQ and PR.

3.7

Notation: 1: 3: 2. 2: or 3. :

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, AC, BD, and angles ∠DAB, ∠ABC, ∠BCD, ∠CDA. Yi Wang Chapter 3. Foundations of Geometry 2 47

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 ;

• 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 3XYZW are congruent under the cor- respondence ABCD ↔ XYZW iff all pairs of corresponding sides and angles under the correspondence are congruent (i.e., CPCF). Such congruence will be denoted by

3ABCD ∼= 3XYZW

. Yi Wang Chapter 3. Foundations of Geometry 2 48

Theorem 3.51 (SASAS Congruence) Suppose that two convex quadrilaterals 3ABCD and 3XYZW satisfy the SASAS Hypothesis under the correspondence ABCD ↔ XYZW . 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 3XYZW . Then 3ABCD ∼= 3XYZW .

Proof: We must prove that the remaining corresponding sides and angles of the two quadri- laterals 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 is a convex quadrilateral having four right angles.

Definition 3.53 Let AB be any , and erect two 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 . 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. Yi Wang Chapter 3. Foundations of Geometry 2 49

Saccheri Quadrilateral in non-Euclidean Geometry

Remark: 1) A Saccheri Quadrilateral in the Poincar`eModel 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 geom- etry. 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. Yi Wang Chapter 3. Foundations of Geometry 2 50

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 () 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. Yi Wang Chapter 3. Foundations of Geometry 2 51

Outline of the proof: (By construction):

Locate the midpoints M and N of sides AB and AC of any triangle ABC, and draw line ←−→ ` = MN. Then drop perpendiculars BB0 and CC0 from B and C to line `. (1) Show 3BCC0B0 (called the Saccheri Quadrilateral associated with ∆ABC) is a Saccheri Quadrilateral. (BB0 = CC0 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 B0C0 equals twice the length of MN. 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 , 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. Yi Wang Chapter 3. Foundations of Geometry 2 52

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 , 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 is the intersection of the circle with a closed half-plane whose passes through the center of the circle, and a major arc of a circle is the intersection of the circle and a central angle and its exterior. We define the measure mACBú of the arc as follows: (see the following figure) Yi Wang Chapter 3. Foundations of Geometry 2 53

Minor Arc Semicircle Major Arc mACBú = m∠AOB mACBú = 180 mACBú = 360 − m∠AOB

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 A1 ∪ A2 = ABC is also an arc. Then m(A1 ∪ A2) = mA1 + mA2.

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: 1) when ABCú is a minor arc or a semicircle. 2) When ABCú is a major arc and BQCú is either a minor arc or a semicircle. 3) When both ABCú and BQCú are major arcs. See the following picture:

Example 3.63 Arc SPTú shown in the following picture is a major arc and is the union of arc SVRú (a minor arc) and arc RPTú (a major arc). Using the angle measures shown in Yi Wang Chapter 3. Foundations of Geometry 2 54 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. 2

Corrolary 3.66 If two tangents PA and PB to a circle O from a common external point −→ P have A and B as the points of contact with the circle, then PA ∼= PB and PO 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 55

2. In general, this is true for any simple closed curves (called Jordan ) .(Jordan Closed Curve Theorem).