Given: ∆ ABC with the internal bisector AD of ∠A which intersects BC at D.

To prove:

First, we construct a line EC ∥ AD which meets BA produced in E.

Now, we have

CE ∥ DA ⇒ ∠2 = ∠3 [Alternate interior angles are equal (transversal AC)]

Also, ∠1 = ∠4 [Corresponding angles are equal (transversal AE)]

We know that AD bisects ∠A ⇒ ∠1 = ∠2

⇒ ∠4 = ∠1 = ∠2 = ∠3

⇒ ∠3 = ∠4

⇒ AE = AC [Sides opposite to equal angles are equal] ……….(i)

Now, consider ∆ BCE,

AD ∥ EC

[By Basic Proportionality theorem]

[∵ BA = AB and AE = AC (From (i))]