Answer :

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,


[By Basic Proportionality theorem]

[ BA = AB and AE = AC (From (i))]

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