Question

$\angle CEG=\angle EAF$.

Answer 1

In $△DBE$,
$DB=BE$
Hence, $\angle DBE=\angle DEB=x$ (I)
In $△DAE$,
$DA=AE$
Hence, $\angle DAE=\angle DEA=y$ (II)
Now, In $△ABE$,
$\angle ABE+\angle BAE+\angle AEB=180$ (Sum of angles of triangle)
$\angle ABE+\angle BAE+\angle BED+\angle DEA=180$
$x+x+y+y=180$
$x+y=90$
$\angle DEB+\angle DEA=90$
$\angle AEB={90}^{\circ }$
Hence, $\angle AEB=\angle AEC={90}^{\circ }$
In $△AEF$,
Sum of angles = 180
$\angle AEF+\angle EAF+\angle EFA=180$
$\angle AEF+\angle EAF+90=180$
$\angle AEF=90-\angle EAF$
We know, $\angle AEC=90$
$\angle AEF+\angle FEC=90$
$90-\angle EAF+\angle FEC=90$
or $\angle EAF=\angle FEC=\angle CEG$