1. reflexive property

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

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

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

5. <1 <2 <3 <4 A B C D Given:AC bisects <A AC bisects <C Prove:∆ABC = ∆ADC 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

6. <1 <2 <3 <4 A B C D Given:AB = AD AC bisects <A Prove:∆ABC = ∆ADC StatementReason AB = AD given AC bisects <A given

7. <1 <2 <3 <4 A B C D Given:AB = AD AC bisects <A Prove:∆ABC = ∆ADC StatementReason AB = AD given AC bisects <A given <1 = <2definition of an angle bisector AC = ACreflexive property ∆ABC = ∆ADCSAS

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

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

10. A B C D <1<2 <3<4 Given:BD bisects <ABC AB = CB Prove:∆ABD = ∆CBD StatementReason BD bisects <ABC given AB = CBgiven

11. A B C D <1<2 <3<4 Given:BD bisects <ABC AB = CB Prove:∆ABD = ∆CBD StatementReason BD bisects <ABC given AB = CBgiven <1 = <2definition of an angle bisector BD = BDreflexive property ∆ABD = ∆CBD SAS

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

13. A B C D <1<2 <3<4 Given:<4 = 90° D is the midpoint of AC Prove:∆ABD = ∆CBD 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