In rectangle $ABCD$,$E$ is the midpoint of $BC$. Prove that $AE = DE$.

(N/A) In rectangle $ABCD$,
$AB = DC$ (Opposite sides of a rectangle are equal)
and $\angle B = \angle C = 90^{\circ}$ (Each angle of a rectangle is $90^{\circ}$).
Also,$E$ is the midpoint of $BC$,so $BE = CE$.
Now,consider $\Delta ABE$ and $\Delta DCE$:
$AB = DC$ (Proved above)
$\angle B = \angle C = 90^{\circ}$ (Proved above)
$BE = CE$ (Given that $E$ is the midpoint of $BC$)
By the $SAS$ (Side-Angle-Side) congruence criterion,$\Delta ABE \cong \Delta DCE$.
Since the triangles are congruent,their corresponding parts are equal.
Therefore,$AE = DE$ (by $CPCT$ - Corresponding Parts of Congruent Triangles).