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EXERCISES 16. Given: Cis the midpoint of AG. HA || GB A Prove: AHAC= ABGC

EXERCISES 16. Given: Cis the midpoint of AG. HA || GB A Prove: AHAC= ABGC

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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4-6 Triangle Congruence: ASA, AAS, and HL EXAMPLES EXERCISES 1 Given: B is the midpoint of AE. ZA = ZE, 16. Given: Cis the midpoint of AG. HA || GB H ZABC = ZEBD Prove: AABC=AEBD Prove: AHAC= ABGC B 17. Given: WX 1 XZ, YZ 1 ZX, WZ = YX Proof: Statements Reasons Prove: AWZX=AYXZ Z, 1. Given 2. Given 3. Given 4. Def. of mdpt. 5. ASA Steps 1, 4, 2 1. ZA = ZE 2. ZABC E ZEBD W 3. B is the mdpt. of AE. 4. AB = EB 5. ΔΑΒC= ΔΕBD 18. Given: ZS and ZV R are right angles. RT = UW. mZT= mZW Prove: ARST :AUVW 4-7 Triangle Congruence: CPCTC EXAMPLES EXERCISES 1 Given: JL and HK bisect each other. 19. Given: M is the midpoint of BD. BC = DC Prove: ZJHG= ZLKG K Prove: 21 = 2 B M D 20. Given: PQ= RQ, PS = RS Prove: QS bisects ZPQR. Proof: Statements Reasons 1. JI and HK bisect 1. Given each other. 2. JG = LG, and HG = KG. 2. Def. of bisect R 3. ZJGH = LLGK 3. Vert. A Thm. 21. Given: H is the midpoint of GJ. L is the midpoint of MK. GM = KJ, GJ = KM, G 4. AJHG = ALKG 5. ZJHG = ZLKG 4. SAS Steps 2, 3 5. СРСТС ZG= ZK Prove: ZGMH= ZKJL M L K 298 Chapter 4 Triangle Congruence
Transcribed Image Text:4-6 Triangle Congruence: ASA, AAS, and HL EXAMPLES EXERCISES 1 Given: B is the midpoint of AE. ZA = ZE, 16. Given: Cis the midpoint of AG. HA || GB H ZABC = ZEBD Prove: AABC=AEBD Prove: AHAC= ABGC B 17. Given: WX 1 XZ, YZ 1 ZX, WZ = YX Proof: Statements Reasons Prove: AWZX=AYXZ Z, 1. Given 2. Given 3. Given 4. Def. of mdpt. 5. ASA Steps 1, 4, 2 1. ZA = ZE 2. ZABC E ZEBD W 3. B is the mdpt. of AE. 4. AB = EB 5. ΔΑΒC= ΔΕBD 18. Given: ZS and ZV R are right angles. RT = UW. mZT= mZW Prove: ARST :AUVW 4-7 Triangle Congruence: CPCTC EXAMPLES EXERCISES 1 Given: JL and HK bisect each other. 19. Given: M is the midpoint of BD. BC = DC Prove: ZJHG= ZLKG K Prove: 21 = 2 B M D 20. Given: PQ= RQ, PS = RS Prove: QS bisects ZPQR. Proof: Statements Reasons 1. JI and HK bisect 1. Given each other. 2. JG = LG, and HG = KG. 2. Def. of bisect R 3. ZJGH = LLGK 3. Vert. A Thm. 21. Given: H is the midpoint of GJ. L is the midpoint of MK. GM = KJ, GJ = KM, G 4. AJHG = ALKG 5. ZJHG = ZLKG 4. SAS Steps 2, 3 5. СРСТС ZG= ZK Prove: ZGMH= ZKJL M L K 298 Chapter 4 Triangle Congruence
4-8 Introduction to Coordinate Proof EXAMPLES EXERCISES 1 Given: ZB is a right angle in isosceles right AABC. E is the midpoint of AB. D is the midpoint of CB. AB = CB Position each figure in the coordinate plane and give the coordinates of each vertex. 22. a right triangle with leg lengths r and s Prove: CE = AD 23. a rectangle with length 2p and width p Proof: Use the coordinates A(0, 2a) , B(0,0), and C(2a, 0). Draw AD and CE. 24. a square with side length 8m For exercises 25 and 26 assign coordinates to each vertex and write a coordinate proof. 25. Given: In rectangle ABCD, E is the midpoint of E AB, Fis the midpoint of BC, Gis the midpoint of CD, and His the midpoint of AD. B D Prove: EF= GH By the Midpoint Formula, (0 +0 2a+0 26. Given: APQR has a right ZQ. Mis the midpoint of PR. E = = (0, a) and 2 (0 + 2a 0+ 0 D= |= (a, 0) Prove: MP = MQ= MR 2 27. Show that a triangle with vertices at (3, 5), (3, 2), and (2, 5) is a right triangle. By the Distance Formula, CE = V(2a – 0)² + (0 – a)² =V4a? + a² = a/5 AD=V(a – 0)² + (0 – 2a)² a² + 4a² = a/5 Thus CE = AD by the definition of congruence. 4-9 Isosceles and Equilateral Triangles EXAMPLE EXERCISES 1 Find the value of x. Find each value. E mZD+ mZE+ mZF= 180° 28. х M K (45 – 3x)° by the Triangle Sum Theorem. mZE = mZF by the Isosceles Triangle Theorem. D 42° mZD+2mZE= 180° Substitution 29. RS R 42 + 2(3x) = 180 Substitute the given values. 2у — 4.5 1.5y 6x = 138 Simplify. x = 23 Divide both sides by 6. 30. Given: AACD is isosceles with ZD as the vertex angle. Bis the midpoint of AC. АВ — х+ 5, ВС3 2х — 3, and CD 3D 2x + 6. Find the perimeter AACD. Study Guide: Review 299
Transcribed Image Text:4-8 Introduction to Coordinate Proof EXAMPLES EXERCISES 1 Given: ZB is a right angle in isosceles right AABC. E is the midpoint of AB. D is the midpoint of CB. AB = CB Position each figure in the coordinate plane and give the coordinates of each vertex. 22. a right triangle with leg lengths r and s Prove: CE = AD 23. a rectangle with length 2p and width p Proof: Use the coordinates A(0, 2a) , B(0,0), and C(2a, 0). Draw AD and CE. 24. a square with side length 8m For exercises 25 and 26 assign coordinates to each vertex and write a coordinate proof. 25. Given: In rectangle ABCD, E is the midpoint of E AB, Fis the midpoint of BC, Gis the midpoint of CD, and His the midpoint of AD. B D Prove: EF= GH By the Midpoint Formula, (0 +0 2a+0 26. Given: APQR has a right ZQ. Mis the midpoint of PR. E = = (0, a) and 2 (0 + 2a 0+ 0 D= |= (a, 0) Prove: MP = MQ= MR 2 27. Show that a triangle with vertices at (3, 5), (3, 2), and (2, 5) is a right triangle. By the Distance Formula, CE = V(2a – 0)² + (0 – a)² =V4a? + a² = a/5 AD=V(a – 0)² + (0 – 2a)² a² + 4a² = a/5 Thus CE = AD by the definition of congruence. 4-9 Isosceles and Equilateral Triangles EXAMPLE EXERCISES 1 Find the value of x. Find each value. E mZD+ mZE+ mZF= 180° 28. х M K (45 – 3x)° by the Triangle Sum Theorem. mZE = mZF by the Isosceles Triangle Theorem. D 42° mZD+2mZE= 180° Substitution 29. RS R 42 + 2(3x) = 180 Substitute the given values. 2у — 4.5 1.5y 6x = 138 Simplify. x = 23 Divide both sides by 6. 30. Given: AACD is isosceles with ZD as the vertex angle. Bis the midpoint of AC. АВ — х+ 5, ВС3 2х — 3, and CD 3D 2x + 6. Find the perimeter AACD. Study Guide: Review 299
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