1. CPCTC

2. <1<2 A B C D E Given:E is the midpoint of BD <B = <D Prove:AB = CD StatementReason E is the midpoint of BDgiven <B = <Dgiven

3. <1<2 A B C D E Given:E is the midpoint of BD <B = <D Prove:AB = CD StatementReason E is the midpoint of BDgiven <B = <Dgiven BE = DEdefinition of a midpoint <1 = <2vertical angle theorem ∆ABC = ∆CDEASA AB = CDCPCTC

4. <1<2 A B C D E Given:E is the midpoint of BD <A = <C Prove:<B = <D StatementReason E is the midpoint of BDgiven <A = <Cgiven

5. <1<2 A B C D E Given:E is the midpoint of BD <A = <C Prove:<B = <D StatementReason E is the midpoint of BDgiven <A = <Cgiven BE = DEdefinition of a midpoint <1 = <2vertical angle theorem ∆ABC = ∆CDEAAS <B = <DCPCTC

6. <1 <2 A B C D E Given:AB // DC AB = DC Prove:BE = DE (using ASA) StatementReason AB // DC given AB = DC given <3<4 <5 <6

7. <1 <2 A B C D E Given:AB // DC AB = DC Prove:BE = DE (using ASA) StatementReason AB // DC given AB = DC given <2 = <6alternate interior angle theorem <1 = <5alternate interior angle theorem ∆ABC = ∆CDEASA BE = DECPCTC <3<4 <5 <6

8. <1 <2 A B C D E Given:AB // DC AB = DC Prove:AE = CE (using AAS) StatementReason AB // DC given AB = DCgiven <3<4 <5 <6

9. <1 <2 A B C D E Given:AB // DC AB = DC Prove:AE = CE (using AAS) StatementReason AB // DC given AB = DC given BE = DEdefinition of a midpoint <1 = <5alternate interior angle theorem <3 = <4vertical angle theorem ∆ABC = ∆CDEAAS AE = CECPCTC <3<4 <5 <6

10. <1 <2 <3 <4 A B C D Given:AB = AD BC = DC Prove:<1 = <2 StatementReason AB = ADgiven BC = DCgiven

11. <1 <2 <3 <4 A B C D Given:AB = AD BC = DC Prove:<1 = <2 StatementReason AB = ADgiven BC = DCgiven AC = ACreflexive property ∆ABC = ∆ADCSSS <1 = <2CPCTC

12. <1 <2 <3 <4 A B C D Given:AC bisects <A AC bisects <C Prove:<B = <D StatementReason AC bisects <A given AC bisects <C given

13. <1 <2 <3 <4 A B C D Given:AC bisects <A AC bisects <C Prove:<B = <D StatementReason AC bisects <A given AC bisects <C given <1 = <2definition of an angle bisector <3 = <4definition of an angle bisector AC = ACreflexive property ∆ABC = ∆ADCASA <B = <DCPCTC

14. A B C D <1<2 <3<4 Given:D is the midpoint of AC AB = CB Prove:<1 = <2 StatementReason D is the midpoint of ACgiven AB = CBgiven

15. A B C D <1<2 <3<4 Given:D is the midpoint of AC AB = CB Prove:<1 = <2 StatementReason D is the midpoint of ACgiven AB = CBgiven AD = CDdefinition of a midpoint BD = BDreflexive property ∆ABD = ∆CBD SSS <1 = <2CPCTC

16. A B C D <1<2 <3<4 Given:<4 = 90° D is the midpoint of AC Prove:<A = <C StatementReason <4 = 90° given D is the midpoint of ACgiven

17. A B C D <1<2 <3<4 Given:<4 = 90° D is the midpoint of AC Prove:<A = <C StatementReason <4 = 90° given D is the midpoint of ACgiven <3 + <4 = 180linear pair postulate <3 + 90 = 180substitution <3 = 90subtraction <3 = <4substitution AD = CDdefinition of a midpoint BD = BDreflexive property ∆ABD = ∆CBD SAS <A = <CCPCTC

18. <1 <2 <3 <4 A B C D Given:AC bisects <A AC bisects <C Prove:<B = <D (w/o using ∆ congruence) StatementReason AC bisects <A given AC bisects <C given

19. <1 <2 <3 <4 A B C D Given:AC bisects <A AC bisects <C Prove:<B = <D (w/o using ∆ congruence) StatementReason AC bisects <A given AC bisects <C given <1 = <2definition of angle bisector <3 = <4definition of angle bisector <1 + <3 + <B= 180triangle sum theorem <2 + <4 + <D = 180triangle sum theorem <1 + <3 + <B = <2 + <4 + <Dsubstitution <2 + <4 + <B = <2 + <4 + <Dsubstitution <B = <Dsubtraction property