Theorem’s We Know 1. Basic Theorems (a) Vertical Angles Theorem: Vertical angles are congruent. (b) Pasch’s Theorem: A line that intersects one side of a triangle intersects one of the other sides. ~ is between AB ~ and AC ~ then AD ~ intersects (c) Crossbar Theorem: If AD BC. (d) External Angle Theorem: An external angle of a triangle is greater than either, opposing, internal angles. 2. Triangle Congruence (a) Angle-Side-Angle: Two triangles with a pair of congruent angles and congruent included side are congruent. (b) Angle-Angle-Side: Two triangles with a pair of congruent angles and a congruent side, not included, are congruent. (c) Isosceles Triangle Theorem: If two sides of a triangle are congruent, the opposite angles are congruent. (d) Isosceles Triangle Theorem Converse: If two angles of a triangle are congruent, the opposite sides are congruent. (e) Isosceles Triangle Theorem Inverse: In a triangle a larger angle is opposite a larger side. 3. More Triangle Theorems (a) Triangle Inequality Theorem: The length of any side of a triangle is less than the sum of the other two side lengths. (b) Points on a perpendicular bisector are equidistant from the endpoints. (c) Points on an angle bisector are equidistant from the sides. (d) The medians of a triangle are concurrent. (e) The perpendicular bisectors of a triangle are congruent. (f) The angle bisectors of a triangle are congruent. 4. Circle Theorems (a) A diameter is longer than any other chord. (b) A diameter that is perpendicular to a chord is the perpendicular bisector. (c) A diameter that bisects a chord is the perpendicular bisector. (d) An inscribed angle is half the measure of the intercepted arc. 1

2

Euclidean Geometry (e) Inscribed angles that intercept the same arc are congruent. (f) Inscribed angles that intercept a diameter are right angles. 5. Parallel Theorems (a) Alternate Interior Angles: If alternate interior angles of a transversal of two lines are congruent, the two lines are parallel. (b) Alternate Interior Angles Converse: If two lines are parallel, the alternate interior angles of a transversal of the two lines are congruent. (c) Triangle Angle Sum: The sum of the interior angles of a triangle is π. (d) The measure of an exterior angle of a triangle equals the sum of the two, opposing, interior angles. (e) Opposing sides of a parallelogram are congruent. (f) If a transversal is cut into congruent segments by parallel lines then any other transversal is also cut into congruent segments.

Chapter 1

Basic Theorems Theorem 1 (Vertical Angles Theorem) Vertical angles are congruent. Proof: Let A − O − B and C − O − D be intersecting lines. We show that ∠AOC ∼ = ∠DOB.

m∠AOB m∠COD π π m∠COB π m∠AOC

= = = = = = =

m∠AOC + m∠COB by postulate 13. m∠COB + m∠BOD by postulate 13. m∠AOC + m∠COB by postulate 14. m∠COB + m∠BOD by postulate 14. (1.1) π − m∠BOD using equation 1.1. (1.2) m∠AOC + (π − m∠BOD) using equations 1.1 and 1.2. m∠BOD.

Thus vertical angles are congruent.



Theorem 2 (Pasch’s Theorem) A line that intersects one side of a triangle intersects one of the other sides. Proof: Consider 4ABC. Let ` be a line that intersects AB. Case 1: ` intersects AC. In this case the theorem is true. Case 2: Suppose ` does not intersect AC. This implies that A and C are on one side of the line by the Plane Separation Postulate. By initial conditions A and B are on different sides of `. Thus by the Plane Separation Postulate ` ∩ BC 6= ∅. Thus a line crossing one side of a triangle intersects one of the other sides.  ~ is between AB ~ and AC ~ then AD ~ Theorem 3 (Crossbar Theorem) If AD intersects BC. 3

4

Euclidean Geometry

~ and AC ~ are Proof: Consider 4ABC and line ` such that A ∈ `. Because AB on opposite sides by the Plane Separation Postulate ` ∩ BC 6= ∅. Thus a line between two sides of a triangle intersects the third side. 

B

E

D

A

C

F

Theorem 4 (External Angle Theorem) An external angle of a triangle is greater than either, opposing, internal angles. Proof: Consider 4ABC. Construct midpoint D of BC. Construct point E such that A − D − E and kADk = kDEk. Let point F be such that A − C − F. We show that external m∠BCF > m∠ABC. Note AD ∼ = DE by construction, ∠ADB ∼ = ∠CDE as vertical angles, and BD ∼ = DC by construction. Thus 4ADB ∼ = 4CDE. Thus ∠ABD ∼ = ∠DCE because CPCTAC. Thus m∠BCF > m∠BCE = m∠ABC. Similarly m∠BCF > m∠BAC. Thus exterior angles are greater than opposite interior angles. 

Chapter 2

Triangle Congruence Theorems Theorem 5 (Angle-Side-Angle) If the vertices of two triangles are in oneto-one correspondence such that two angles and the included side of one triangle are congruent, respectively, to two angles and the included side of the second triangle, then the triangles are congruent. ∼ ∠F DE, Proof: Suppose that triangles 4ABC and 4DEF have ∠CAB = ∠CBA ∼ = ∠F ED, and AB ∼ = DE. Case 1: If AC ∼ = DF or CB ∼ = EF , then the triangles are congruent by SAS. Case 2: WoLOG (without loss of generality) suppose kACk > kDF k. Thus there exists a point C 0 such that A − C 0 − C and AC 0 ∼ = DF . Thus by SAS 4ABC 0 ∼ 6 = 4DEF. In particular ∠ABC 0 ∼ = ∠DEF. But, ∠DEF ∼ = ∠ABC ∼ = 0 ∼ ∠ABC . Thus the supposition kACk > kDF k is absurd, and AC = DF . Therefore if two angles and the included side are congruent, the triangles are congruent.  Theorem 6 (Angle-Angle-Side) If the vertices of two triangles are in oneto-one correspondence such that two angles and the side opposite one of them in one triangle are congruent to the corresponding parts of the second triangle, then the triangles are congruent. Proof: Suppose triangles 4ABC and 4DEF have ∠CAB ∼ = ∠F DE, ∠CBA ∼ = ∼ ∠F ED, and AC = DF . Case 1: If AB ∼ = DE, then the triangles are congruent by SAS. Case 2: WoLOG suppose kABk > kDEk. Thus there exists a point B 0 such that A − B 0 − B and AB 0 ∼ = DE. Thus by SAS 4AB 0 C ∼ = 4DEF. In particular 0 ∠AB 0 C ∼ ∠DEF. Thus ∠AB C ∼ = = ∠DEF ∼ = ∠ABC. But ∠AB 0 C is an exterior angle to 4CB 0 B, which implies ∠AB 0 C > ∠ABC. Thus the supposition kABk > kDEk is absurd, and AB ∼ = DE. Therefore if two angles and one of the sides not between them are congruent, the triangles are congruent.  5

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Euclidean Geometry

Theorem 7 (Isosceles Triangle Theorem) If two sides of a triangle are congruent, then the angles opposite those sides are congruent. Proof: Suppose that 4P QR has P Q ∼ = RQ. Construct the bisector of ∠P QR. By the Crossbar theorem this angle bisector intersects P R at a point T such that P − T − R. Note ∠P QT ∼ = QT by identity. = ∠RQT, because P T is an angle bisector. QT ∼ Therefore 4QP T ∼ 4QRT by SAS. Because CPCTAC, ∠QP T ∼ = = ∠QRT. Thus angles opposite congruent sides of a triangle are congruent. 

Chapter 3

More Triangle Theorems Theorem 8 If a point is equidistant from the endpoints of a line segment, then it is on the perpendicular bisector of that line segment Proof: Suppose the point P is equidistant from the endpoints of AB. Construct the bisector of ∠AP B. By the Crossbar theorem this angle bisector intersects AB in a point T such that A − T − B. AP ∼ = BP by construction. ∠AP T ∼ = ∠BP T by construction and P T ∼ = PT ∼ by identity. By SAS 4AP T ∼ 4BP T. In particular ∠AT P ∠BT P, be= = cause CPCTAC. These are also supplementary. Thus m∠AT P = m∠BT P and m∠AT P +m∠BT P = π. Solving these simultaneous equations gives m∠AT P = m∠BT P = π/2. Hence P T is the perpendicular bisector of AB. Therefore the set of points are equidistant  Theorem 9 (Inverse of Isosceles Triangle Theorem) If two sides of a triangle are not congruent, then the angles opposite those sides are not congruent, and the larger angle is opposite the larger side. Proof: Consider 4ABC with kACk > kBCk. Because kACk > kBCk, there exists a point D on CB such that C − D − B and CD ∼ = AC. By the isosceles triangle theorem ∠CAD ∼ = ∠CDA. By definition ∠CBA is an exterior angle of 4ABD. As a result m∠CBA > m∠CDA by the exterior angle theorem. Further, because B is interior to ∠CAD, m∠CAD > m∠CAB by postulate. Thus m∠CBA > m∠CAB. Thus the larger angle is opposite the larger side in a triangle. 

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Euclidean Geometry

Chapter 4

Circle Theorems Theorem 10 A chord that is not a diameter is shorter than a diameter. Proof: Consider a circle with center O and chord CD which is not a diameter. Construct a diameter AB. Also construct OC and OD. Consider 4OCD. By the triangle inequality theorem, kOCk + kODk > kCDk. Note that OC, OD are diameters. Thus kABk = kOCk + kODk > kCDk. Thus a diameter is a maximum chord.  Theorem 11 A diameter orthogonal to a chord is the perpendicular bisector of the chord. Proof: Consider a circle with center O. Suppose diameter AB is orthogonal to chord CD. Case 1: CD is a diameter. Thus CD ∩ AB = O by definition of diameter and (SMSG) postulate 1. OC and OD are radii and thus congruent. Thus AB bisects CD. Case 2: CD is not a diameter. Let CD ∩ AB = P. Note ∠CP O, ∠DP O are right angles. OC ∼ = OD by construction, and OP ∼ = OP by reflexivity. Because the hypotenuses and legs are congruent, the right triangles are congruent. Thus CP = DP , and AB is the perpendicular bisector of CD.  Theorem 12 A line tangent to a circle is orthogonal to the radius at the point of tangency. Proof: Consider a circle with center O. Let ` be a line tangent to the circle at point P. We will prove this theorem by contradiction. Suppose ` is not orthogonal to radius OP . Construct point Q on ` such that OQ ⊥ `. Construct point R such that P −Q−R and RQ ∼ = QP . Note OQ ∼ = OQ by reflexivity, and ∠OQP ∼ = ∠OQR by construction. Thus by ∼ SAS 4OQR ∼ OR 4OQP, and by CPCTAC = OP . Because OP is a radius, = OR must be a radius and R must also be on the circle. But this is absurd, because ` is tangent at P. Thus OP ⊥ `.  9

10

Euclidean Geometry

Theorem 13 The measure of an inscribed angle is one half the measure of the intercepted arc. Proof: Consider a circle with center O, and inscribed ∠CAB. Case 1: Suppose AB is a diameter. Note m∠COB +m∠COA = π by postulates 13 and 14. Note CO ∼ = OA, because they are both radii. Also m∠CAO + m∠ACO + m∠COA m∠CAO + m∠CAO + m∠COA 2m∠CAO + m∠COA 2m∠CAO + (π − m∠COB) 2m∠CAO m∠CAO

= = = = =

π. π. π. π. m∠COB. 1 = m∠COB. 2

Case 2: Suppose AB is not a diameter. Construct diameter AD. By case 1 m∠CAD = 12 m∠COD. Also by case 1 m∠DAB = 12 m∠DOB. Now by the angle addition postulates m∠CAB

= m∠CAD + m∠DAB 1 1 = m∠COD + m∠DOB 2 2 1 = (m∠COD + m∠DOB) 2 1 m∠COB. = 2

Thus the measure of the inscribed angle is half the measure of the intercepted arc. 

Chapter 5

Parallel Theorems Theorem 14 (Triangle Angle Sum) The sum of the interior angles of a triangle is π. Proof: Consider 4ABC. Construct line ` such that B ∈ ` and ` k AC. Let P − B − R be points on `. Note m∠P BA + m∠ABC + m∠CBR = π by postulates 13 and 14. By the Alternate Interior Angle Theorem ∠BCA ∼ = ∠CBR and ∠BAC ∼ = ∠P BA. Thus m∠BAC + m∠ABC + m∠BCA = π. Thus the sum of the interior angles of a triangle is π.  Theorem 15 The measure of an exterior angle of a triangle equals the sum of the two, opposing, interior angles. Proof: Consider 4ABC. Let P be a point such that A − C − P. Note m∠BCP + m∠ACB = π by postulate 14. m∠CAB + m∠ABC + m∠ACB = π by angle sum theorem. m∠CAB + m∠ABC + (π − m∠BCP ) = π. m∠CAB + m∠ABC = m∠BCP. Thus the measure of an exterior angle equals the sum of the measures of the opposing, interior angles.  Theorem 16 Opposing sides of a parallelogram are congruent. Proof: Let ABCD be a parallelogram. Construct AC. By the alternating, interior angle theorem ∠BCA ∼ = ∠CAD and ∠BAC ∼ = ∼ AC by identity. Thus 4ABC ∠DCA. Also AC ∼ = = 4CDA by ASA. Thus AB ∼ = DC and BC ∼ = AD by CPCTAC. Thus opposite sides of a parallelogram are congruent.  11

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Euclidean Geometry

Theorem 17 If a transversal is cut into congruent segments by parallel lines then any other transversal is also cut into congruent segments. Proof: Let `, m, n be three, parallel lines. Let t1 be a transversal such that t1 ∩ n = A, t1 ∩ m = B, t1 ∩ ` = C, and AB ∼ = BC. Consider transversal t2 . Let t2 ∩ n = A0 , t2 ∩ m = B 0 , t2 ∩ ` = C 0 . Construct transversal tp such that tp k t1 and tp ∩ m = B 0 . Let tp ∩ n = D and tp ∩ ` = E. By construction BB 0 k AD and AB k DB 0 . Thus ABB 0 D is a parallelogram. Similarly BCEB 0 is a parallelogram. Thus by previous theorem AB ∼ = DB 0 and BC ∼ = B 0 E. Further AB ∼ = BC is given, so DB 0 ∼ = B 0 E by transitivity. Note ∠C 0 B 0 E ∼ = ∠A0 B 0 D by vertical angle theorem. Note also that ∠C 0 EB 0 ∼ = 0 0 ∠B DA , because they are alternating, interior angles. Thus 4C 0 B 0 E ∼ = 4A0 B 0 D. Finally C 0 B 0 ∼ = B 0 A0 by CPCTAC. Therefore if a transversal is cut into congruent segments by parallel lines then any other transversal is also cut into congruent segments.