$17 \cdot 100$ In $\triangle A B C$, $B C=a, A C=b, A B=c, \angle C=90^{\circ} . C D$ and $B E$ are two medians of $\triangle A B C$, and $C D \perp B E$. Then $a: b: c=$
(A) $1: 2: 3$.
(B) $3: 2: 1$.
(C) $\sqrt{3}: \sqrt{2}: 1$.
(D) $1: \sqrt{2}: \sqrt{3}$.
(Anhui Province, China Junior High School Mathematics Competition, 1997)
[Solution] As shown in the figure, let $C D$ intersect $B E$ at $F$, and let $E F=x, D F=y$, then $B F=2 x$, $C F=2 y$.
By the properties of similar triangles, we have
$$
\begin{array}{l} 
a^{2}=B E \cdot B F=6 x^{2},\left(\frac{b}{2}\right)^{2}=E F \cdot E B=3 x^{2},\left(\frac{c}{2}\right)^{2}=y^{2}+4 x^{2},
\end{array}
$$

which means $a^{2}=6 x^{2}, b^{2}=12 x^{2}, c^{2}=4 y^{2}+16 x^{2}$.
Also, $C F^{2}=E F \cdot F B$, which is $4 y^{2}=2 x^{2}$, then $c^{2}=18 x^{2}$, $\therefore a^{2}: b^{2}: c^{2}=1: 2: 3$, i.e., $a: b: c=1: \sqrt{2}: \sqrt{3}$. Therefore, the answer is $(D)$.