Question

In the given figure, bisector of $\angle$BAC intersects side BC at point D. Prove that AB > BD

Answer 1

Given: AD is the bisector of ∠A
∴ ∠DAB = ∠DAC
Now, ∠BDA is the exterior angle of the ∆DAC
∴ ∠BDA > ∠DAC
⇒ ∠BDA > ∠DAB (∵ ∠DAC = ∠DAB)
⇒ AB > BD (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence proved.