In a convex pentagon ABCDE: $AB=AC$, $AE=AD$, $\angle CAD= \angle ABE + \angle AEB$, M is the midpoint of BE. Prove $2AM=CD$

Question

I'd love to know how to prove this. I don't exactly know how to begin.

I can recognize the two isosceles triangles ABC and ADE.

I can also express M as $\frac{B+E}2$.

But then I'm stuck...

Answer 1

Let $F$ is placed on the ray $BA$ such that $A$ is a mid-point of $BF$.

Thus, $AM=\frac{1}{2}FE$.

But since $$\measuredangle FAE=\measuredangle ABE+\measuredangle AEB=\measuredangle CAD,$$ we obtain: $\Delta FAE\cong\Delta CAD,$ which gives $FE=CD$ and we are done!

Answer 2

Using formula of median’s length we have $$4AM^2 =2AB^2+2AE^2-EB^2$$ By cosine law in $\triangle CAD$ and using the fact we have two isosceles triangles $$CD^2 =AB^2+AE^2-2ABAE\cos \angle CAD$$ If we substitute $\angle CAD = \angle ABE + \angle AEB = \pi - \angle BAE$ $$CD^2 =AB^2+AE^2+2ABAE\cos \angle BAE$$ By cosine law in $\triangle BAE$ $$2ABAE\cos \angle BAE = AB^2 + AE^2 - EB^2$$ Substituting this result in previous equation we have $$4AM^2 = CD^2$$