Question

In ∆ABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB = seg AC, then prove that ED || BC.

Answer 1

In ∆ABC, ray BD bisects ∠ABC. [Given] 

∴ AB/BC = AE/EB  (i) [Property of angle bisector of a triangle] 

Also, in ∆ABC, ray CE bisects ∠ACB. [Given] 

∴ AC/BC = AE/EB  (ii) [Property of angle bisector of a triangle] 

But, seg AB = seg AC (iii) [Given] 

∴ AB =  BC (iv) [From (ii) and (iii)] 

∴ AD/DC = AE/EB  [From (i) and (iv)] 

∴ ED || BC [Converse of basic proportionality theorem]