18.77 In the square $ABCD$ (as shown in the figure), the side length is 12 inches. A segment $AE$ is drawn, with $E$ on side $DC$, and $DE$ is 5 inches long. The perpendicular bisector of $AE$ intersects $AE$, $AD$, and $BC$ at points $M$, $P$, and $Q$, respectively. The ratio of segments $PM$ and $MQ$ is
(A) $5: 12$.
(B) $5: 13$.
(C) $5: 19$.
(D) $1: 4$.
(E) $5: 21$.
(23rd American High School Mathematics Examination, 1972)
[Solution] As shown in the figure, draw a line through $M$ parallel to $AB$, intersecting $AD$ at $R$ and $BC$ at $S$. Then,
$$
R M=\frac{1}{2} D E=2 \frac{1}{2} \text { inches. }
$$

Therefore, the length of $MS$ is $12-2 \frac{1}{2}=9 \frac{1}{2}$ inches.
$$
\begin{aligned}
\because \quad \text { Rt } \triangle P M R & \sim \mathrm{Rt} \triangle Q M S, \\
\therefore \quad P M: M Q & =R M: M S \\
& =2 \frac{1}{2}: 9 \frac{1}{2} \\
& =5: 19 .
\end{aligned}
$$

Therefore, the answer is $(C)$.