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.

We present another proof of the fact that $E$ lies on line $I H$. Since all five internal bisectors of $A B C D E$ meet at $I$, this pentagon has an inscribed circle with center $I$. Let this circle touch side $B C$ at $T$.

Applying Brianchon's theorem to the (degenerate) hexagon $A B T C D E$ we conclude that $A C, B D$ and $E T$ are concurrent, so point $E$ also lies on line $I H T$, completing the proof.

We present yet another proof that $E I \perp B C$. In pentagon $A B C D E$, $\angle E < 180^{\circ} \Longleftrightarrow \angle A + \angle B + \angle C + \angle D > 360^{\circ}$. Then $\angle A + \angle B = \angle C + \angle D > 180^{\circ}$, so rays $E A$ and $C B$ meet at a point $P$, and rays $B C$ and $E D$ meet at a point $Q$. Now,
$$\angle P B A = 180^{\circ} - \angle B = 180^{\circ} - \angle D = \angle Q D C$$
and, similarly, $\angle P A B = \angle Q C D$. Since $A B = C D$, the triangles $P A B$ and $Q C D$ are congruent with the same orientation. Moreover, $P Q E$ is isosceles with $E P = E Q$.

In Solution 1 we have proved that triangles $I A B$ and $I C D$ are also congruent with the same orientation. Then we conclude that quadrilaterals $P B I A$ and $Q D I C$ are congruent, which implies $I P = I Q$. Then $E I$ is the perpendicular bisector of $P Q$ and, therefore, $E I \perp P Q \Longleftrightarrow E I \perp B C$.