Proving equal angle in regular pentagon

Question

showing that all the angles in a regular pentagon are the same euclid style.

In this other page, How can one of the answer get to the conclusion of BED + ABE = ABC

This is my first time using, sorry if there is some rules I don't understand

Answer 1

Let $m\angle ABE=m\angle BAC=m\angle AEB=x$. Let $P$ be the intersection of $BE$ and $AC$. We see that $\triangle APE \cong \triangle BPC$ (ASA). Thus, $EP=PC$ so $\triangle EPC$ is isosceles. Moreover, $\triangle EPC$ and $\triangle APB$ are similar ($\angle APB \cong \angle EPC$ as vertical and both triangles are isosceles).

From the above we conclude that $m\angle PEC = m\angle ECP=x$.

We already showed that $m\angle BED = \pi - m\angle ABC $ (from the answer you are referring to). Using that $ m\angle ABC = \pi - 2 m\angle AEB$ we conclude that $m\angle BED=2x$ and that $m\angle CED=m\angle BED-m\angle PEC=x$.

Finally, we can conclude that $m\angle EDC=\pi-2x=m\angle ABC$ ($\triangle EDC$ is isosceles so $\angle CED \cong \angle CDE$)