Question

Prove that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

Answer 1

Let the triangle be ABC with AD as the bisector of ∠A which meets BC at D. 

We have to prove: 

BD/DC = AB/AC

Draw CE || DA, meeting BA produced at E. 

CE || DA 

Therefore,

∠2 = ∠3 

(alternate angles) and ∠1 = ∠4 (corresponding angles) 

But, 

∠1 = ∠2 

Therefore, 

∠3 = ∠4 

⇒ AE = AC 

In ∆BCE, DA || CE. 

Applying Thales’ theorem, we gave: 

BD/DC = AB/AE

⇒ BD/DC = AB/AC 

This completes the proof.