Show that if a convex pentagon is equilateral, and two of its adjacent inner angles measure $108$ degrees, then the pentagon is equiangular.
My Attempt. Consider the figure below.
Let $\angle CDE = \angle BCD = 180$ degrees. Currently, I don't really have a good idea on where to start, but my initial attempt was to show that I can draw a pentagram and show that $\triangle ABC, \triangle BCD, \triangle CDE, \triangle DEF,\triangle EFA$ are congruent and hence, the figure is equiangular. It is rather easy to show that $\triangle BCD$ and $\triangle CDE$ are congruent by SAS congruence, however, I am unable to show congruences with other triangles. I feel like I am on the wrong tract. Any suggestions?
Look at $BCDE$. One can complete this quadrilateral into a regular pentagon $A'BCDE$. All one has to do is prove that $A'=A$. But the triangles $ABE$ and $A'BE$ both have the same side-lengths, and so are congruent. As both $A$ and $A'$ lie above the line $BE$ then $A=A'$.