7.28 In $\triangle A B C$, $\angle C$ is a right angle, $A C=3$ and $B C=4$. In $\triangle A B D$, $\angle A$ is a right angle, and $A D=12$. Points $C$ and $D$ are on opposite sides of $A B$, and the line through $D$ parallel to $A C$ intersects $C B$ at $E$. If $\frac{D E}{D B}=\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers, then $m+n$ equals
(A) 25.
(B) 128.
(C) 153.
(D) 243.
(E) 256.
(42nd American High School Mathematics Examination, 1991)
[Solution] Let $\angle ABC = \alpha, \angle ABD = \beta, \angle DBE = \gamma$.
From the given information, we have
$$
AB = \sqrt{AC^2 + BC^2} = \sqrt{3^2 + 4^2} = 5,
$$

and
$$
BD = \sqrt{AD^2 + AB^2} = \sqrt{12^2 + 5^2} = 13.
$$

Then
$$
\begin{aligned}
\sin \gamma & = \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\
& = \frac{3}{5} \cdot \frac{5}{13} + \frac{4}{5} \cdot \frac{12}{13} = \frac{63}{65},
\end{aligned}
$$

and
$$
\sin \gamma = \frac{DE}{DB} = \frac{m}{n} = \frac{63}{65},
$$

Therefore, the answer is $(B)$.