Prove that ED=EF

Question

In the diagram below $AD=DC$ and $AE=EB$ and both triangles $AEB$ and $ADC$ are right angel triangles and $M$ is the midpoint of $BC$ also $MD=MF$. Prove that $ED=EF$

Answer 1

Join $BF$.

$\triangle BMF\cong \triangle CMD$ by SAS, so $$\begin{align*} BF&= CD\\ \angle MBF &= \angle MCD\\ &= \angle MCA + 45^\circ \end{align*}$$

Consider angles around point $B$,

$$\begin{align*} \angle EBF &= 360^\circ - \angle EBA - \angle ABM - \angle MBF\\ &= 360^\circ - 45^\circ - \angle ABM - (\angle MCA + 45^\circ)\\ &= 270^\circ - \angle ABM -\angle MCA\\ &= 45^\circ + (180^\circ -\angle ABM - \angle MCA) + 45^\circ\\ &= \angle EAB + \angle BAC + \angle CAD\\ &= \angle EAD \end{align*}$$

So $\triangle EAD\cong \triangle EBF$ by SAS, and

$$ED=EF$$

Answer 2

Let $\mathcal{R}$ be a rotation around $D$ for $90^{\circ}$ and let $\mathcal{Z}$ a reflection across $M$. Then we see that composition $\mathcal{J}= \mathcal{Z}\circ \mathcal{R}$ of these two transformation is also rotation for $270^{\circ} =-90^{\circ}$ around some point $X$. Now since $$ \mathcal{J}: A\mapsto C\mapsto B$$ we see that $E$ is a center of this new rotation $\mathcal{J}$. But $$ \mathcal{J}: D\mapsto D\mapsto F$$ we see it take $D$ to $F$ and so $EF = ED$ (and $\angle FED = 90^{\circ}$)