Throughout the solution, we refer to $\angle A, \angle B, \angle C, \angle D$, and $\angle E$ as internal angles of the pentagon $A B C D E$. Let the perpendicular bisectors of $A C$ and $B D$, which pass respectively through $B$ and $C$, meet at point $I$. Then $B D \perp C I$ and, similarly, $A C \perp B I$. Hence $A C$ and $B D$ meet at the orthocenter $H$ of the triangle $B I C$, and $I H \perp B C$. It remains to prove that $E$ lies on the line $I H$ or, equivalently, $E I \perp B C$. Lines $I B$ and $I C$ bisect $\angle B$ and $\angle C$, respectively. Since $I A=I C, I B=I D$, and $A B=$ $B C=C D$, the triangles $I A B, I C B$ and $I C D$ are congruent. Hence $\angle I A B=\angle I C B=$ $\angle C / 2=\angle A / 2$, so the line $I A$ bisects $\angle A$. Similarly, the line $I D$ bisects $\angle D$. Finally, the line $I E$ bisects $\angle E$ because $I$ lies on all the other four internal bisectors of the angles of the pentagon. The sum of the internal angles in a pentagon is $540^{\circ}$, so
$$\angle E=540^{\circ}-2 \angle A+2 \angle B .$$
In quadrilateral $A B I E$,
$$\begin{aligned} \angle B I E & =360^{\circ}-\angle E A B-\angle A B I-\angle A E I=360^{\circ}-\angle A-\frac{1}{2} \angle B-\frac{1}{2} \angle E \\ & =360^{\circ}-\angle A-\frac{1}{2} \angle B-\left(270^{\circ}-\angle A-\angle B\right) \\ & =90^{\circ}+\frac{1}{2} \angle B=90^{\circ}+\angle I B C, \end{aligned}$$
which means that $E I \perp B C$, completing the proof. !