5. 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 $a: b: c=(\quad)$.
(A) $1: 2: 3$
(B) $3: 2: 1$
(C) $\sqrt{3}: \sqrt{2}: 1$
(D) $1: \sqrt{2}: \sqrt{3}$
5. (D).

As shown in the figure, let $C D$ and $B E$ intersect at $F$. Suppose $E F=x$, $D F=y$, then $B F=2 x$, $C F=2 y$. By the properties of similar triangles, we have $a^{2}=B E \cdot B F=6 x^{2},\left(\frac{b}{2}\right)^{2}=E F \cdot E B=3 x^{2},\left(\frac{c}{2}\right)^{2}=y^{2}+4 x^{2}$, which means $a^{2}=6 x^{2}, b^{2}=12 x^{2}, c^{2}=4 y^{2}+16 x^{2}$. 
Also, $C F^{2}=E F \cdot F B, 4 y^{2}=2 x^{2}$, thus $c^{2}=18 x^{2}$, $\therefore a^{2}: b^{2}: c^{2}=1: 2: 3$, which means $a: b: c=1: \sqrt{2}: \sqrt{3}$.