In the proof, we will use $∠A, ∠B, ∠C, ∠D, ∠E$ to denote the interior angles of the pentagon $ABCDE$. Let the point $I$ be the intersection of the perpendicular bisectors of segment $AC$ and segment $BD$. Note that the perpendicular bisector of $AC$ passes through point $B$, and the perpendicular bisector of $BD$ passes through point $C$. Hence $BD \perp CI$ and $AC \perp BI$. Therefore, the intersection of $AC$ and $BD$ is the orthocenter $H$ of triangle $BIC$, and $IH \perp BC$. It remains to prove that the point $E$ lies on the line $IH$, that is, $EI \perp BC$.

The lines $IB$ and $IC$ bisect $∠B$ and $∠C$ respectively. Since $IA = IC$, $IB = ID$, and $AB = BC = CD$, the three triangles $IAB$, $ICB$, $ICD$ are all congruent. This gives
$$∠IAB = ∠ICB = \frac{∠C}{2} = \frac{∠A}{2},$$

hence $IA$ bisects $\angle A$. Similarly, $ID$ bisects $\angle D$. Finally, $IE$ also bisects $\angle E$, because $I$ lies on the angle bisectors of the other four interior angles of the pentagon.

The sum of the interior angles of a convex pentagon is $540^\circ$, so in the quadrilateral $ABIE$,
$$\begin{aligned} \angle BIE &= 360^\circ - \angle EAB - \angle ABI - \angle AEI \\ &= 360^\circ - \angle A - \frac{1}{2}\angle B - \frac{1}{2}\angle E \\ &= 360^\circ - \angle A - \frac{1}{2}\angle B - (270^\circ - \angle A - \angle B) \\ &= 90^\circ + \frac{1}{2}\angle B = 90^\circ + \angle IBC, \end{aligned}$$
hence $EI \perp BC$. This completes the proof.