State and prove the converse of angle bisector theorem.
                               

Given: ABC is a Δ ; AD divides BC in the ratio of the sides containing the angles ∠A to meet BC at D.
i.e. 
To prove: AD bisects ∠A.
Construction: Draw CE || DA to meet BA produced at E.
Proof: In Δ ABC, CE || DA cut by AE.
∠BAD = ∠AEC (corresponding angle) ----(i)
Similarly CE || DA cut by AC
∠ DAC = ∠ACE (alternate angles) ----(ii)
In DBEC; CE || AD
AB = BD (BPT)
AE    DC
But AB = BD (given)
      AC    DC
AB = AB
AE     AC
AE = AC
⇒  ∠AEC =  ∠ACE (isosceles property) ----(iii)
According to equation (i), (ii) and (iii)  ∠BAD = ∠DAC
⇒ AD bisects  ∠A.