$17 \cdot 101$ The lengths of the sides of a certain triangle are $13, 14, 15$, the orthocenter is $H, AD$ is the altitude to the side of length 14, then $H D: H A$ equals
(A) 3:11.
(B) $5: 11$.
(C) $1: 2$.
(D) $2: 3$.
(E) $25: 33$.
(15th American High School Mathematics Examination, 1964)
［Solution］In the triangle, the lengths of the segments are set as shown in the figure. By the Pythagorean theorem, we have
$$
\begin{aligned}
14^{2}-q^{2}=B E^{2} & =13^{2}-p^{2}, \\
q^{2}-p^{2} & =27 .
\end{aligned}
$$

Also, $p+q=15$, which gives $q=\frac{42}{5}, p=\frac{33}{5}$.
From $13^{2}-r^{2}=A D^{2}=15^{2}-s^{2}$,
we get $s^{2}-r^{2}=56$, and $s+r=14$,
which gives $r=5, s=9$.
By $A D^{2}=13^{2}-5^{2}$, we get $A D=12$;
and $B E^{2}=13^{2}-\left(\frac{33}{5}\right)^{2}$, we get $B E=\frac{56}{5}$.
$\because \triangle H D B \sim \triangle H E A \quad \therefore \quad \frac{t}{u}=\frac{r}{p}=\frac{\frac{56}{5}-u}{12-t}=\frac{H B}{H A}$,
and $r=5, p=\frac{33}{5}, u=\frac{33}{25} t$, substituting these into the above equation gives $t=\frac{435}{116}$.
Thus, $12-t=\frac{957}{116}$. Therefore, $H D: H A=435: 957=5: 11$. Hence, the answer is (B).