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Given that ∠ADB and ∠CDB are right angles, and BD¯¯¯¯¯ bisects ∠B , prove that D is the midpoint of AC¯¯¯¯¯ . The figure shows triangle A B C. Point D lies on side A C. 1. ∠ADB and ∠CDB are right angles (given)2. BD¯¯¯¯¯ bisects ∠B (given)3. ∠ADB≅∠CDB (Right ∠≅ Thm.)4. BD¯¯¯¯¯≅DB¯¯¯¯¯ (Reflex. Prop. of ≅)5. ∠DBA≅∠DBC (Def. of ∠ bisector) 6. △ABD≅ △CBD (ASA, Steps 3, 4, 5)7. AD¯¯¯¯¯≅DC¯¯¯¯¯ (CPCTC)8. AD=DC (Def of ≅)9. D is midpoint of AC¯¯¯¯¯ (definition of midpt) 1. ∠ADB and ∠CDB are right angles (given)2. BD¯¯¯¯¯ bisects ∠B (given)3. ∠ADB≅∠CDB (Right ∠≅ Thm.)4. ∠DBA≅∠DBC (Def. of ∠ bisector)5. AB¯¯¯¯¯≅CB¯¯¯¯¯ (∠ Bisector Thm.)6. △ABD≅ △CBD (ASA, Steps 3, 4, 5)7. AD¯¯¯¯¯≅DC¯¯¯¯¯ (CPCTC)8. AD=DC (Def of ≅)9. D is midpoint of AC¯¯¯¯¯ (definition of midpt) 1. ∠ADB and ∠CDB are right angles (given)2. BD¯¯¯¯¯ bisects ∠B (given)3. ∠ADB≅∠CDB (Right ∠≅ Thm.)4. ∠DBA≅∠DBC (Def. of ∠ bisector)5. ∠DAB≅∠DCB (Interior ∠ Thm.) 6. △ABD≅ △CBD (AAA, Steps 3, 4, 5)7. AD¯¯¯¯¯≅DC¯¯¯¯¯ (CPCTC)8. AD=DC (Def of ≅)9. D is midpoint of AC¯¯¯¯¯ (definition of midpt) 1. ∠ADB and ∠CDB are right angles (given)2. BD¯¯¯¯¯ bisects ∠B (given)3. ∠ADB≅∠CDB (Right ∠≅ Thm.)4. BD¯¯¯¯¯≅DB¯¯¯¯¯ (Reflex. Prop. of ≅)5. AD¯¯¯¯¯≅DC¯¯¯¯¯ (Def. of ∠ bisector)6. △ABD≅ △CBD (SAS, Steps 3, 4, 5)7. AD=DC(CPCTC)8. D is midpoint of AC¯¯¯¯¯ (definition of midpt)

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