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ABC is a triangle in which ∠B = 2 ∠C . D is a point on BC such that AD bisects ∠BAC and AB = CD. Prove that ∠BAC = 72°.

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Given that ABC is a triangle in which ∠B = 2 ∠C and D is a point on BC such that AD bisectors ∠BAC and AB = CD

We have to Prove that ∠BAC = 72°

Now, draw the angular bisector of ABC, which meets AC in P. join PD

So, by SAS congruence criterion, we have ΔABP = ΔDCP

Now,

∠BAP = CDF and

AP = DP [Corresponding parts of congruent triangles are equal]

From the above two equations, we get

We know,

Sum of angles in a triangle = 180°

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