8. In the cyclic quadrilateral $A B C D$, $\angle A=$
$$
\begin{array}{l}
\angle 52.5^{\circ}, \angle B=97.5^{\circ}, \\
\angle A O B=120^{\circ}, A B=a, \\
B C=b, C D=c, D A=d .
\end{array}
$$

Then the area $S$ of this quadrilateral is ( ).
(A) $\frac{1}{2}(a d+b c)$
(B) $\frac{1}{2}(a b+c d)$
(C) $\frac{1}{2}(a c+b d)$
(D) $\frac{1}{4}(a d+b c+a b+c d)$
8. C

On $\overparen{D C B}$, construct $\overparen{D C^{\prime}}=$
$\overparen{B C}$, it is easy to prove that $D C^{\prime}=b, B C^{\prime}$
$=c, \triangle A D C^{\prime}, \triangle A B C^{\prime}$
are right triangles, and
$\triangle B C D^{\prime}$ is equal in area to $\triangle D C^{\prime} B$, so we should choose C. Below, we only prove $\angle A B C^{\prime}=90^{\circ}$. In fact,
$$
\begin{array}{l}
\angle B O C=\angle A O C-\angle A O B \\
=2\left(180^{\circ}-97.5^{\circ}\right)-120^{\circ}=45^{\circ}, \\
\angle B O C^{\prime}=\angle D O C \\
=\angle D O B-\angle B O C \\
=2 \times 52.5^{\circ}-45^{\circ}=60^{\circ} .
\end{array}
$$
$\therefore \triangle O B C^{\prime}$ is an equilateral triangle, $\angle A B C^{\prime}$
$$
=\angle A B O+\angle B O C^{\prime}=30^{\circ}+60^{\circ}=90^{\circ} \text {. }
$$