$7 \cdot 47$ In $\triangle A B C$, $\angle C=90^{\circ}, \angle A$'s bisector intersects $B C$ at $D$, then $\frac{A B-A C}{C D}$ equals.
(A) $\sin A$.
(B) $\cos A$.
(C) $\operatorname{tg} A$.
(D) $\operatorname{ctg} A$.
(China Guangzhou, Wuhan, Fuzhou and other five cities junior high school mathematics league, 1985)