18.26 For any convex quadrilateral $ABCD$ with diagonals intersecting at $O$, let the areas of $\triangle ABO$, $\triangle BCO$, $\triangle CDO$, $\triangle DAO$ be $S_{1}$, $S_{2}$, $S_{3}$, $S_{4}$, respectively. Which of the following conclusions must be true?
(A) $S_{1}+S_{3}=S_{2}+S_{4}$.
(B) $S_{1}-S_{3}=S_{2}-S_{4}$.
(C) $S_{1} \cdot S_{3}=S_{2} \cdot S_{4}$.
(D) $S_{1}: S_{3}=S_{2}: S_{4}$.
［Solution］From the formula for the area of a triangle, we have
$$
\begin{array}{ll}
S_{1}=\frac{1}{2} A O \cdot B O \cdot \sin \angle A O B, & S_{2}=\frac{1}{2} B O \cdot C O \cdot \sin \angle B O C, \\
S_{3}=\frac{1}{2} C O \cdot D O \cdot \sin \angle C O D, & S_{4}=\frac{1}{2} D O \cdot A O \cdot \sin \angle A O D .
\end{array}
$$

Thus, \( S_{1} S_{3}=\frac{1}{4} A O \cdot B O \cdot C O \cdot D O \cdot \sin \angle A O B \cdot \sin \angle C O D \),
and \( S_{2} S_{4}=\frac{1}{4} A O \cdot B O \cdot C O \cdot D O \cdot \sin \angle B O C \cdot \sin \angle A O D \).
Given that \( \angle A O B + \angle B O C = 180^{\circ} \) and \( \angle C O D + \angle A O D = 180^{\circ} \),

we have \( \sin \angle A O B = \sin \angle B O C \) and \( \sin \angle C O D = \sin \angle A O D \).
Therefore, \( S_{1} S_{3} = S_{2} S_{4} \).
Hence, the correct choice is \((C)\).