$7 \cdot 47$ 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)
[Solution] As shown in the figure, draw $D E \perp A B$. It is easy to prove that $\triangle A C D \cong \triangle A E D$.
From $A C=A E$, we know $E B=A B-A C$.
And $\angle E D B=\angle A$.
In the right triangle $\triangle D E B$, we have
$$
\frac{E B}{E D}=\operatorname{tg} \angle E D B=\operatorname{tg} A .
$$

Therefore, the answer is $(C)$.