If $AB$ and $CD$ are perpendicular to $BC$ and $AB = CD$, show that $EB = EC.$ A B C E D

1. We are given that $\angle ABE = \angle DCE = 90^ \circ$ and $AB = CD$.
We need to find the value of $EB$.
2. In $\triangle ABE$ and $\triangle DCE$, we have \begin{aligned} &AB = CD &&\text{[Given]} \\ &\angle ABE = \angle DCE &&\text{[Each 90}^ \circ] \\ &\angle AEB = \angle CED &&\text{[Vertically opposite angles]} \\ &\therefore \space \triangle ABE \cong \triangle DCE && \text{[By AAS Criterion]} \\ \end{aligned}
3. As corresponding parts of congruent triangles are equal, we have $\bf {EB = EC}$.