17-82 In $\triangle A B C$, $\angle C=90^{\circ}$, and $C D$ is its angle bisector. Given $A D=p$, $D B=q$, then the altitude $h$ from $A B$ is
(A) $\frac{p q}{p^{2}+q^{2}}$.
(B) $\frac{p+q}{p^{2}+q^{2}}$.
(C) $\frac{p q(p+q)}{p^{2}+q^{2}}$.
(D) $\frac{p^{2}+q^{2}}{p q(p+q)}$.
(Chinese Junior High School Mathematics Competition, Hangzhou, Zhejiang Province, 1985)
[Solution] Given that $CD$ bisects $\angle ACB$, then $\frac{AC}{BC}=\frac{AD}{BD}=\frac{p}{q}$, let $AC=pt, BC=qt$.
By the Pythagorean theorem, we have $(pt)^{2}+(qt)^{2}=(p+q)^{2}$,
thus $\quad t^{2}=\frac{(p+q)^{2}}{p^{2}+q^{2}}$.
Also, $2 S_{\triangle ABC}=AC \cdot BC=h \cdot AB$,

then $h=\frac{AC \cdot BC}{AB}=\frac{pt \cdot qt}{p+q}=\frac{pq t^{2}}{p+q}=\frac{pq(p+q)}{p^{2}+q^{2}}$.
Therefore, the answer is $(C)$.