Consider $\Delta DAE$ and $\Delta DCG$,
$\begin{array}{ll}
DA=DC & \text{(sides of square)} \\
AE=CD & \text{(given)} \\
\angle DAE = \angle DCG = 90^\circ & \text{(properties of square)}
\end{array}$
$\therefore \Delta DAE \cong \Delta DCG \text{ (S.A.S.)}$.
Therefore, we have $\angle ADE = \angle CDG = 25^\circ$ (corr. $\angle$s, $\cong \Delta$s).
Hence, we have
$\begin{array}{rcl}
\angle EDF & = & 90^\circ – \angle ADE – \angle FDC \\
& = & 90^\circ – 25^\circ – 20^\circ \\
& = & 45^\circ
\end{array}$
Also, $DE = DG$ (corr. sides, $\cong \Delta$s).
Note that
$\begin{array}{rcl}
\angle FDG & = & \angle CDG + \angle CDF \\
& = & 20^\circ + 25^\circ \\
& = & 45^\circ
\end{array}$
Consider $\Delta DEF$ and $\Delta DGF$,
$\begin{array}{ll}
DE = DG & \text{(given)} \\
DF = DF & \text{(common side)} \\
\angle EDF=\angle GDF=45^\circ & \text{(given)}
\end{array}$
$\therefore \Delta DEF \cong \Delta DGF \text{ (S.A.S.)}$.
Therefore, $\angle DFE = \angle DFG$ (corr. $\angle$s, $\cong \Delta$s).
Hence, we have
$\begin{array}{rcl}
\angle DFE & = & \angle DFG \\
& = & 180^\circ – \angle CDF – \angle DCF \\
& = & 180^\circ – 90^\circ – 20^\circ \\
& = & 70^\circ
\end{array}$
