show that the mid-perpendicular segments AB and CD intersect at the bisector

Question

The ABCDE pentagon has all angles of the same measure. How to show that the mid-perpendicular segments AB and CD intersect at the bisector of the angle DEA? Is this pentagon regular or another? Maybe this problem has simple solution, but i don,t konow how to do it.

My attempts:

Let E be a cut point AB i CD. I've tried notice in triangles AEF and DEF, but im stuck.

Answer 1

Let $F$ be an intersection point of the mid-perpendiculars, $\measuredangle FAB=\measuredangle FBA=\alpha,$ $\measuredangle FCD=\measuredangle FDC=\beta$ and $\measuredangle FEA=x$.

Thus, by law of sines we obtain: $$1=\frac{AF}{EF}\cdot\frac{BF}{AF}\cdot\frac{CF}{BF}\cdot\frac{DF}{CF}\cdot\frac{EF}{DF}=$$ $$=\frac{\sin{x}}{\sin(108^{\circ}-\alpha)}\cdot1\cdot\frac{\sin(108^{\circ}-\alpha)}{\sin(108^{\circ}-\beta)}\cdot1\cdot\frac{\sin(108^{\circ}-\beta)}{\sin(108^{\circ}-x)}=\frac{\sin{x}}{\sin(108^{\circ}-x)},$$ which gives $x=54^{\circ}$ and we are done!

Another way (L.Radzivilovski,R.Leker).

Let $AE\cap BC=\{P\}$ and $DE\cap BC=\{Q\}$.

Thus, inside bisectors from $P$ and $Q$ of $\Delta PQE$ are placed on mid-perpendiculars to segments $AB$ and $CD$ respectively.