$7 \cdot 46 \triangle ABC$ In, $\angle C=90^{\circ}$, and $CB > CA$, point $D$ is on $BC$ and $\angle CAD=2 \angle DAB$. If $\frac{AC}{AD}=\frac{2}{3}$, and $\frac{CD}{BD}=\frac{m}{n}$, where $m, n$ are coprime positive integers, then $m+n=$
(A) 10.
(B) 14.
(C) 18.
(D) 22.
(E) 26.
(49th American High School Mathematics Examination, 1998)
[Solution] Let $\angle B A D=\alpha$, then $\angle D A C=2 \alpha$,
$$
\begin{array}{l}
\angle B A C=3 \alpha \text{. } \\
\frac{C D}{B D}=\frac{S_{\triangle A C D}}{S_{\triangle} A B D}=\frac{A C \sin 2 \alpha}{A B \sin \alpha}=\frac{A C}{A B} 2 \cos \alpha \\
=2 \cos 3 \alpha \cos \alpha=8 \cos ^{4} \alpha-6 \cos ^{2} \alpha \text{. } \\
\therefore \quad \cos ^{2} \alpha=\frac{5}{6} \text{. } \\
\end{array}
$$
Since $\cos 2 \alpha=\frac{A C}{A D}=\frac{2}{3}$, that is $2 \cos ^{2} \alpha-1=\frac{2}{3}$,

then $\frac{C D}{B D}=8 \cdot\left(\frac{5}{6}\right)^{2}-6 \cdot\left(\frac{5}{6}\right)=\frac{5}{9}=\frac{m}{n}$, so $m+n=14$. Therefore, the answer is $(B)$.