Given in Δ ABC, AD bisects the ∠A meeting BC at D

BD = DC and ∠BAD = ∠ CAD…………. 1

Construction:- Extend BA to E and join C to E such CE ∥ AD……… 4

∠BAD = ∠AEC (corresponding ∠s)……………… 2

∠CAD = ∠ ACE (alternate interior ∠s)……………….. 3

From 1 , 2 and 3

∠ ACE = ∠AEC

In Δ AEC

∠ ACE = ∠AEC

∴ AC = AE (sides opposite to equal angles are equal)……….. 5

In Δ BEC

AD ∥ CE (From ….4)

And D is midpoint of BC (given)

By converse of midpoint theorem

A line drawn from the midpoint of a side, parallel to the opposite side of the triangle meets the third side in its middle and is half of it

∴ A is midpoint of BE

BA = AE……… 6

From 5 and 6

AB = BC

⇒ ΔABC is an isosceles triangle