Let $O$ be the centre of the square. Then $O$ is the incentre of $\triangle AEZ$, due to the bisectors of $\angle EAZ$ and $\angle EZA$.

Therefore $\angle OEZ=\frac12 \angle AEZ=45^\circ$.

$\triangle OEF\sim\triangle CED$ (note that $|OF|=\frac12|CD|$ and $OF\perp DC$), therefore $\angle OEC=90^\circ$ and so $\angle ZEC=45^\circ$.

Let $O$ be the centre of the square. Then $O$ is the incentre of $\triangle AEZ$, due to the bisectors of $\angle EAZ$ and $\angle EZA$.

Therefore $\angle OEZ=\frac12 \angle AEZ=45^\circ$.

$\triangle OEF\sim\triangle CED$ (note that $|OF|=\frac12|CD|$ and $OF\perp DC$), therefore $\angle OEC=90^\circ$ and so $\angle ZEC=45^\circ$.

Another proof is that $\triangle DEC\sim\triangle DFZ$ (drop the perpendicular from $E$ to $CD$). As $\angle ZFC=45^\circ$, so is $\angle ZEC$.

Let $O$ be the centre of the square. Then $O$ is the incentre of $\triangle AEZ$, due to the bisectors of $\angle EAZ$ and $\angle EZA$.

Therefore $\angle OEZ=\frac12 \angle AEZ=45^\circ$.

$\triangle OEF\sim\triangle CED$ (note that $|OF|=\frac12|CD|$), therefore $\angle OEC=90^\circ$ and so $\angle ZEC=45^\circ$.

Let $O$ be the centre of the square. Then $O$ is the incentre of $\triangle AEZ$, due to the bisectors of $\angle EAZ$ and $\angle EZA$.

Therefore $\angle OEZ=\frac12 \angle AEZ=45^\circ$.

$\triangle OEF\sim\triangle CED$ (note that $|OF|=\frac12|CD|$ and $OF\perp DC$), therefore $\angle OEC=90^\circ$ and so $\angle ZEC=45^\circ$.