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Prove that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

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Prove that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

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Let the triangle be ABC with AD as the bisector of ∠A which meets BC at D. 

We have to prove:

BD/DC = AB/AC

This completes the proof.

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