1. 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) $\operatorname{ctg} A$
-1 (C).
As shown in the figure, draw $D E \perp A P$ at $F$. Since $D$ is on the angle bisector of $\angle A$, we have $D E$
$$
\begin{array}{l}
=C D . A E=A C \\
\quad \therefore \frac{A B-A C}{C D}=\frac{A B-A E}{D E} \\
\quad=\frac{B E}{D E}=\operatorname{tg} \angle B D E=\operatorname{tg} A .
\end{array}
$$