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,


AD EC


[By Basic Proportionality theorem]


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


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