21. [12] Let $A B C D$ be a quadrilateral inscribed in a circle with center $O$. Let $P$ denote the intersection of $A C$ and $B D$. Let $M$ and $N$ denote the midpoints of $A D$ and $B C$. If $A P=1, B P=3, D P=\sqrt{3}$, and $A C$ is perpendicular to $B D$, find the area of triangle $M O N$.
Answer: $\frac{3}{4}$
We first prove that $O N P M$ is a parallelogram. Note that $A P D$ and $B P C$ are both $30^{\circ}-60^{\circ}-90^{\circ}$ triangles. Let $M^{\prime}$ denote the intersection of $M P$ and $B C$. Since $\angle B P M^{\prime}=\angle M P D=30^{\circ}$, we have $M P \perp B C$. Since $O N$ is the perpendicular bisector of $B C$, we have $M P / / N O$. Similarly, we have $M O / / N P$. Thus $O N P M$ is a parallelogram. It follows that the area of triangle $M O N$ is equal to the area of triangle $M P N$, which is equal to $\frac{1}{2} \cdot 1 \cdot 3 \cdot \sin \angle M P N=\frac{1}{2} \cdot 1 \cdot 3 \cdot \sin 150^{\circ}=\frac{3}{4}$.