Proving question on triangles.

Question

I have a question that seems very difficult to solve by myself:

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

Please help.

Answer 1

Let me try.

Let $BE$ be a bisector of $\angle B$. One then has $\triangle ABE \sim \triangle ACD$. So, $$\frac{AB}{AC} = \frac{AE}{AD} = \frac{BE}{CD}.$$

Moreover, one has $$\frac{AE}{AD} = \frac{AB}{AB+BD}.$$

We get $AC = AB + BD = CD + BD = BC$. So, $\angle BAC = \angle ABC = 2\angle BCA$. Finally, we get $\angle BAC = \frac{2\times 180^\circ}{5} = 72^\circ.$