17・188 As shown in the figure, in the right triangle $\triangle ABC$, $\angle C = 90^{\circ}$, and $CD$ is the altitude on the hypotenuse $AB$. Points $E$ and $F$ are taken on $BC$ and $CA$ respectively, such that $\triangle FED$ is similar to $\triangle ABC$. The number of such $\triangle FED$ is
(A) 1.
(B) 2.
(C) 3.
(D) More than 3.
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1991)
[Solution] Clearly, as long as $\angle E D F=90^{\circ}$, from the fact that points $F, D, E, C$ are concyclic, we can prove
$$
\angle E F D=\angle B C D \text {. }
$$

And since $C D \perp A B$, we know $\angle B C D=\angle C A B$, thus we have
$$
\mathrm{Rt} \triangle F D E \backsim \mathrm{R} \triangle A C B \text {. }
$$

Therefore, there are infinitely many $\triangle F E D$ similar to $\triangle A B C$.
Hence, the answer is $(D)$.