Example 2. As shown in the figure, in the isosceles right $\triangle ABC$, $AD$ is the median on the right side $BC$, $BE \perp AD$, and intersects $AC$ at $E$, $EF \perp BC$. If $AB = BC = a$, then $EF$ equals ( ).
(A) $\frac{1}{3} a$
(B) $\frac{1}{2}-2$
(C) $\frac{2}{3} a$
(D) $\frac{2}{5} a$ O100, Xi'an Junior High School Mathematics Competition
Draw $CG$ perpendicular to $BC$ intersecting the extension of $BE$ at $G$.
There is
$\triangle A B D \cong \triangle B C G$.

So $C G=\frac{a}{2}$.
That is $\frac{1}{a}+\frac{1}{\frac{a}{2}}=\frac{1}{E F}$. Thus $E F=\frac{1}{3} a$. Therefore, the answer is (A).