$17 \cdot 17$ In $\triangle A B C$, $\angle A C B$ $=90^{\circ}, C D$ and $C E$ are the altitude and the angle bisector of $\angle C$ on the base $A B$, respectively. If $\triangle C E D \sim \triangle A B C$. Then $\angle E C D$ equals
(A) $18^{\circ}$.
(B) $20^{\circ}$.
(C) $22.5^{\circ}$.
(D) $30^{\circ}$.
[Solution] Given $\angle A C B=90^{\circ}$, and $C D \perp A B$, also $\triangle C E D \sim \triangle A B C$, then $\angle C E D=\angle A B C$, so $\triangle C E B$ is an isosceles triangle.
Thus, $C D$ bisects $\angle E C B$. Also, $C E$ bisects $\angle A C B$, so $\angle E C D=\frac{1}{2} \angle E C B=\frac{1}{2} \cdot \frac{1}{2} \angle A C B=22.5^{\circ}$. Therefore, the answer is $(C)$.