$7 \cdot 33$ In an isosceles triangle, the sum of the altitudes from the two equal sides equals the altitude from the base, then the sine value of the base angle is
(A) $\frac{1}{2}$.
(B) $\frac{\sqrt{3}}{2}$.
(C) $\frac{\sqrt{15}}{4}$.
(D) $\frac{1}{4}$.
(Shenyang City, Liaoning Province, "Yucai Cup" Junior High School Mathematics Competition, 1988)
[Solution] As shown in the figure, let $AB = AC = m$, then $AD = m \sin C$, and $CD = m \cos C$, $BC = 2m \cos C$.

Also, $BE = BC \sin C = 2m \sin C \cos C$.
According to the problem, $BE + CF = AD$, and $BE = CF$,
$$
\begin{array}{c}
\therefore \quad 4m \sin C \cos C = m \sin C. \\
\text { That is } \sin C(4 \cos C - 1) = 0.
\end{array}
$$
$\because C$ is an acute angle,
$\therefore \sin C \neq 0$, hence $\quad \cos C = \frac{1}{4}$.
$$
\therefore \sin C = \sqrt{1 - \cos^2 C} = \sqrt{1 - \left(\frac{1}{4}\right)^2} = \frac{\sqrt{15}}{4}.
$$

Therefore, the answer is $(C)$.