Solution:

Because of the given assumptions, $\triangle ABC$ and $\triangle BCD$ are isosceles. Hence the perpendicular bisector of $AC$ passes through $B$ and the perpendicular bisector of $BD$ passes through $C$. Both intersect at the point $I$ (see figure).

Since $BD \perp CI$ and $AC \perp BI$, $AC$ and $BD$ intersect at the orthocenter of the triangle $BCI$, and it follows that $IH \perp BC$. If we can now show that $EI \perp BC$, the claim follows, because through $I$ there can be only one line orthogonal to $BC$. Since $BI$ and $CI$ are also the angle bisectors of $\angle CBA$ and $\angle DCB$ respectively, it follows that $\overline{IA}=\overline{IC}$ as well as $\overline{IB}=\overline{ID}$. Because of $\overline{AB}=\overline{BC}=\overline{CD}$, the triangles $ABI$, $BCI$ and $CDI$ are congruent. Therefore $\angle BAI=\angle ICB=\frac{1}{2} \angle DCB=\frac{1}{2} \angle BAE$, so that $IA$ is the angle bisector of $\angle BAE$. Analogously it holds that $ID$ is the angle bisector of $\angle EDC$.

The composition of the reflections in the axes $AI, BI, CI, DI$ and $EI$ is a reflection in an axis with the fixed points $E$ and $I$. Hence $I$ also lies on the angle bisector of $\angle AED$.

Now $\angle AED=540^\circ-2 \angle CBA-2 \angle BAE$ holds, and thus in the quadrilateral $ABIE$ it follows that:
$$\angle BIE=360^\circ-\angle BAE-\angle IBA-\angle AEI=360^\circ-\angle BAE-\frac{1}{2} \angle CBA-\left(270^\circ-\angle CBA-\angle BAE\right)$$
$$=90^\circ+\frac{1}{2} \angle CBA=90^\circ+\angle CBI.$$
By the exterior angle theorem it follows that $EI \perp BC$. ㅁ.