I is true. Since $ABCD$ is a rectangle, then $AB\text{//}DC$. Hence, we have
$\begin{array}{rcll}
\angle ABE & = & \angle DGF & \text{(corr. $\angle$s, $AB$//$DC$)} \\
\end{array}$
In $\Delta ABE$,
$\begin{array}{rcll}
\angle AEB & = & 90^\circ & \text{(given)} \\
\angle ABE & = & 180^\circ -\angle AEB- \angle BAE & \text{($\angle$ sum of $\Delta$)} \\
\angle ABE & = & 180^\circ -90^\circ -\angle BAE & \\
\angle ABE & = & 90^\circ -\angle BAE
\end{array}$
Hence, $\angle DGF = 90^\circ -\angle BAE$.
On the other hand, since $ABCD$ is a rectangle, then we have
$\begin{array}{rcll}
\angle BAD & = & 90^\circ & \text{(properties of rectangle)} \\
\angle DAE & = & \angle BAD -\angle BAE & \\
\angle DAE & = & 90^\circ -\angle BAE & \\
\angle DAE & = & \angle ABE & \\
\angle DAE & = & \angle DGF
\end{array}$
II is true. In $\Delta BCE$ and $\Delta CGE$,
$\begin{array}{rcllll}
\angle BEC & = & \angle CEG & = & 90^\circ & \text{(given)} \\
\end{array}$
Also,
$\begin{array}{rcll}
\angle ABC & = & 90^\circ & \text{(properties of rectangle)} \\
\angle CBE & = & 90^\circ -\angle ABE \\
\angle ABE & = & \angle CGE & \text{(alt. $\angle$s, $AB$//$DC$)} \\
\angle GCE & = & 180^\circ -\angle CEG -\angle CGE & \text{($\angle$ sum of $\Delta$)} \\
\angle GCE & = & 180^\circ -90^\circ -\angle CGE & \\
\angle GCE & = & 90^\circ -\angle CGE \\
\angle GCE & = & 90^\circ -\angle ABE & \text{(proved)} \\
\angle DCE & = & \angle CBE & \text{(proved)}
\end{array}$
Hence, $\Delta BCE \sim \Delta CGE$ (A.A.).
III is true. In $\Delta BCE$ and $\Delta FCE$,
$\begin{array}{rcllll}
\angle CEF & = & \angle CEB & = & 90^\circ \\
CE & = & CE & & & \text{(common side)} \\
CF & = & AD & & & \text{(given)} \\
AD & = & CB & & &\text{(properties of rectangle)} \\
CF & = & CB \\
\end{array}$
Hence, $\Delta BCE \cong \Delta FCE$ (R.H.S.).
