$20 \cdot 74$ Let $AB$ be a chord of $\odot O$, and $CD$ be a diameter of $\odot O$ that intersects the chord $AB$. Let $M=|S_{\triangle CAB}-S_{\triangle DAB}|, N=2S_{\triangle OAB}$, then
(A) $M>N$.
(B) $M=N$.
(C) $M<N$.
(D) The relationship between $M$ and $N$ is uncertain.
(China Junior High School Mathematics League, 1995)
[Solution] As shown in the figure, draw $D E \perp A B, C F \perp A B$ (feet of the perpendiculars are $E, F$), and let the extension of $C F$ intersect $\odot O$ at $G$. Connect $D G$.

Since $\angle D G C=90^{\circ}$ (as $C D$ is the diameter, hence $E F C D$ is a rectangle, thus $D E=G F$.

Assume $C F>F G$. Draw $O H \perp C G$ at $H$, and draw $O L \perp A B$ at $L$. Then $H$ is the midpoint of $C G$, and
$$
\begin{array}{l}
C F-F G=C H+H F-F G \\
=2 H F=2 O L \text {. } \\
M=\left|S_{\triangle C A B}-S_{\triangle D A B}\right| \\
=\frac{1}{2} A B \cdot C F-\frac{1}{2} A B \cdot D E \\
=\frac{1}{2} A B(C F-F G)=A B \cdot O L=2 S_{\triangle O A B} \text {. } \\
\end{array}
$$

Therefore, the answer is $(B)$.