4. As shown in Figure 3, from a known

point $A$ on the circle, two
(A) $\frac{S_{1}}{S_{2}} \geqslant \frac{a}{b}$
(c) $\frac{S_{1}}{S_{2}}=\frac{a}{b}$
(C) $\frac{S_{1}}{S_{2}}>\frac{a}{b}$
(D) $\frac{S_{1}}{S_{2}} \leqslant \frac{a}{b}$

chords $A B$ and $A C$ are drawn with lengths $a$ and $b$ respectively, and $a<b$. Chord $A D$ bisects $\angle B A C$, and the areas of figures $A B D$ and $A C D$ are $S_{1}$ and $S_{2}$ respectively. Then the relationship between $\frac{S_{1}}{S_{2}}$ and $\frac{a}{b}$ is
4.C.

Connect $B D, D C, \angle B A D=\angle C A D$, we know, $\frac{S_{\triangle A B D}}{S_{\triangle A C D}}=\frac{a}{b}$. We can set $S_{\triangle \triangle B D}=a k, S_{\triangle A C D}=b k$.
It is easy to prove $S_{3}$ lipD $=S_{\text {Зemm. }}$ (set as $S_{0}$ ), then we have
$$
\begin{array}{l}
S_{1}=a k+S_{0}, S_{2}=b k+S_{0}, \\
\frac{S_{1}}{S_{2}}-\frac{a}{b}=\frac{a k+S_{0}}{b k+S_{0}}-\frac{a}{b}=\frac{S_{0}(b-a)}{b\left(b k+S_{0}\right)}>0 \\
(\because a\frac{a}{b}$.
$$