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Using Two Pairs of Congruent Triangles in a Proof. AEB, AC AD, 1. Given BC BD. 1). 2. AB AB 2. Reflexive. C. 3. ACB ADB 3. SSS SSS. 4. CAE DAE 4. CPCTC. E. B. A. 5. AE AE 5. Reflexive. 6. ACE ADE 6. SAS SAS. D. 7. CE DE 7. CPCTC.
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AEB, AC AD, 1. Given • BC BD 1) 2. AB AB 2. Reflexive C 3.ACB ADB 3. SSS SSS 4. CAE DAE 4. CPCTC E B A 5. AE AE 5. Reflexive 6. ACE ADE 6. SAS SAS D 7.CE DE 7. CPCTC Prove: CE DE
2) Plan: • NK NV, 1. Given • IK IV E K V 2. IN IN 2. Reflexive 3. KIN VIN 3. SSS SSS 4. <KNE <VNE 4. CPCTC I 5. NE NE 5. Reflexive 6. KEN VEN 6. SAS SAS • KE VE 7. CPCTC • NE bisects KV 8. A segment is bisected if it is cut in half. N Prove: NE bisects KV
E D G A F B • AD CB, DC BA 1. Given • EF bisects BD 3) • BD BD 2. Reflexive • ABD CDB 3. SSS SSS C 4. CDB ABD 4. CPCTC 5. DG BG 5. A bisector cuts in half 6.DGE BGF 6. Vertical Angles are congruent 7. DGE BGF 7. ASA ASA 8. FG EG 8. CPCTC Prove: FG EG
1 2, AP CP 1. Given • PQ, PAB, PCD, • AQD, CQB B D • PQ PQ 2. Reflexive • PAQ PCQ 3. SAS SAS Q • QA QC 4. CPCTC • QAP QCP A C 5. BAQ is suppl to QAP 5. Linear pairs are DCQ is suppl. to PCQ supplements. 6. BAQ DCQ 6. It 2 <‘s are , then their suppelments are 1 2 P 7. BQA DQC 7. Vertical <‘s are Prove: QB QD 8. PAQ PCQ 8. ASA ASA 9. QB QD 9. CPCTC
Homework: Pairs of Congruent Triangles • AC and BD bisect 1. Given • each other at G, • EGF 1) D E C 2. AG CG, BG DG 2. A bisector cuts in half 3. AGB CGD 3. Vertical ’s are G 4. AGB CGD 4. SAS SAS 5. A C 5. CPCTC 6.AGF CGE 6. Vertical ’s are A 7. AGB CGD 7. ASA ASA F B 8. GE GF 8. CPCTC Prove: GE GF
AC AD, BC BD 1. Given • AB intersects CD • at E 2) • AB AB 2. Reflexive • ABC ABD 3. SSS SSS C 4. CAB DAB 4. CPCTC 5. ACE ADE 5. If 2 sides of a are , then opposite ’s are 2 A B E 1 6.ADE ACE 6. ASA ASA 7. 1 2 2. CPCTC D Prove: 1 2
3) • RP RQ, SP SQ 1. Given R 2. RS RS 2. Reflexive 3. RPS RQS 3. SSS SSS 4. PRS QRS 4. CPCTC 5. RPT RQT 5. If 2 sides of a are , then opposite ’s are 6. RPT RQT 6. ASA ASA 7. PT QT 7. CPCTC S • RT bisects PQ 8. A segment is bisected if its cut in half P Q T Prove: RT bisects PQ