Example 5. For an isosceles trapezoid $\triangle B C D$ with a base angle of $67.5^{\circ}$, a circle is drawn with the leg $B C$ as the diameter, intersecting the lower base $A B$ at $E$. 11. It is exactly tangent to the leg $A D$ at $M$. Find $B E: A E$. (1981, Beijing High School Competition)
As shown in Figure 5, let the center be \( O \), and \( OE, OA \). Then, by the Angle Bisector Theorem, we have
\[
\begin{aligned}
& \frac{B E}{A E} \\
= & \frac{O B \cdot \sin \angle B O E}{O A \cdot \sin \angle E O A}
\end{aligned}
\]

Since \( O E = O B \) and \( \angle E B O = 67.5^{\circ} \), it follows that \( \angle B O E = 45^{\circ} \), and \( O E \parallel D A \), thus \( \angle E O A = \angle M A O \). Next, we find \( O M \). Since \( M \) is the point of tangency, we have
\[
\begin{array}{l}
O A \cdot \sin \angle E O A = O A \cdot \sin \angle M A O \\
\quad = O M = O B .
\end{array}
\]

Therefore, \(\frac{B E}{A E} = \sin \angle B O E = \sin 45^{\circ} = \frac{\sqrt{2}}{2}\).