$17 \cdot 58$ In the isosceles $\mathrm{Rt} \triangle A B C$, $A D$ is the median on the right-angled side $B C$, and $B E \perp A D$ intersects $A C$ at $E, E F \perp B C$. If $A B=B C=a$, then $E F$ equals
(A) $\frac{1}{3} a$.
(B) $\frac{1}{2} a$.
(C) $\frac{2}{3} a$.
(D) $\frac{2}{5} a$.
［Solution］It is easy to prove $\triangle A B D \backsim \triangle B F E$, then $\frac{A B}{B D}=\frac{B F}{F E}$. Also, by the given condition, $\angle C=45^{\circ}$, so $E F=F C$, thus
$$
\frac{B F}{F C}=\frac{B F}{F E}=\frac{A B}{B D}=2 \text {, }
$$
then
$$
E F=F C=\frac{a}{3} .
$$

Therefore, the answer is $(A)$.