In the adjoining figure, bisector of ∠B AC intersects side BC at poi

1.	In the adjoining figure, bisector of ∠B AC intersects side BC at point D. Prove that AB > BD.Given: Bisector of ∠BAC intersects side BC at point D. To prove: AB > BD
Answer» ∠BAD ≅ ∠DAC ….(i) [Seg AD bisects ∠BAC] ∠ADB is the exterior angle of ∆ADC. ∴ ∠ADB > ∠DAC ….(ii) [Property of exterior angle] ∴ ∠ADB > ∠BAD ….(iii) [From (i) and (ii)] In AABD, ∠ADB > ∠BAD [From (iii)] ∴ AB > BD [Side opposite to greater angle is greater]

In the adjoining figure, bisector of ∠B AC intersects side BC at point D. Prove that AB > BD.Given: Bisector of ∠BAC intersects side BC at point D. To prove: AB > BD

Answer»

∠BAD ≅ ∠DAC ….(i) [Seg AD bisects ∠BAC]

∠ADB is the exterior angle of ∆ADC.

∴ ∠ADB > ∠DAC ….(ii) [Property of exterior angle]

∴ ∠ADB > ∠BAD ….(iii) [From (i) and (ii)]

In AABD,

∠ADB > ∠BAD [From (iii)]

∴ AB > BD [Side opposite to greater angle is greater]