As shown in Figure 2, in the square $A B C D$, the diagonals $A C$ and $B D$ intersect at

point $O$, point $F$ is on side $C D$,
the extension of $A F$ intersects the
extension of $B C$ at
point $E$, the extension of $O F$
intersects $D E$ at point $M$. Find the measure of $\angle O M D$.
From the given conditions, we have
$$
\begin{array}{l}
A B=A D, \angle A D F=\angle A B E=90^{\circ}, \\
A D \parallel B E \Rightarrow \angle D A F=\angle B E A .
\end{array}
$$

Therefore, $\triangle A D F \sim \triangle E B A \Rightarrow \frac{A D}{B E}=\frac{D F}{A B}$
$$
\Rightarrow B E \cdot D F=A D \cdot A B=A D^{2} \text {. }
$$

Also, $O D \cdot B D=A D^{2}$, so
$$
B E \cdot D F=O D \cdot B D \Rightarrow \frac{B E}{O D}=\frac{B D}{D F} \text {. }
$$

Since $\angle D B E=45^{\circ}=\angle O D F$, we have
$\triangle D B E \sim \triangle F D O$
$$
\Rightarrow \angle D F O=\angle B D E \text {. }
$$

Thus, $\angle O M D=\angle D F O-\angle F D M$
$$
=\angle B D E-\angle F D M=\angle B D C=45^{\circ} \text {. }
$$