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°