$17 \cdot 83$ In $\triangle A B C$, $B C=3, A C=4$. Also, $A E$ and $B D$ are the medians to sides $B C$ and $A C$, respectively, and $A E \perp B D$. Then the length of $A B$ is
(A) $\sqrt{5}$.
(B) $3 \sqrt{6}$.
(C) $2 \sqrt{2}$.
(D) $\sqrt{7}$.
(E) cannot be determined.
(Chongqing, Sichuan Province, China Junior High School Mathematics Competition, 1983)
[Solution] Let $A E=3 x, B D=3 y$, then $A G=2 x, G E=x, B G=2 y$, $G D=y$.
Also, $A D=\frac{1}{2} A C=2, \quad B E=\frac{1}{2} B C=\frac{3}{2}$.
Thus, $(2 x)^{2}+y^{2}=A D^{2}=4$, and $x^{2}+(2 y)^{2}=B E^{2}=\frac{9}{4}$,
Adding the two equations gives $5 x^{2}+5 y^{2}=\frac{25}{4}$, which simplifies to $x^{2}+y^{2}=\frac{5}{4}$, then $4 x^{2}+4 y^{2}=5$, i.e., $A G^{2}+B G^{2}=5$.
Therefore, $A B^{2}=A G^{2}+B G^{2}=5$, so $A B=\sqrt{5}$. Hence, the answer is $(A)$.