Angle bisectors of isosceles trapezoid $\square ABCD$ meet at $P$. If $|AP|=3$ and $\angle APD=120^\circ$, what is the area of the trapezoid?

Question

Let $\square ABCD$ be an isosceles trapezoid where $AD$ is a base. Let point $P$ be a point inside the trapezoid such that $PA$, $PB$, $PC$, $PD$ are angle bisectors of $A$, $B$, $C$, $D$. If $|PA| = 3$ and $\angle APD = 120^\circ$, then what is the area of the trapezoid?

My try: I got $|AD| = {3}\sqrt3$ using the cosine law. I then connected the midpoint of $BC$ to $A$ and $D$, respectively, to get a equilateral triangle. I got the length of $AD$ and the height of the trapezoid, but can't seem to get the length of $BC$.

Answer 1

Hint: $$A=\frac{1}{2}\cdot 9\sin(120^{\circ})+2\cdot 3 \cdot \frac{\tan(30^{\circ})}{2}+\frac{1}{2}(3\tan(30^{ \circ}))^2\sin(60^{\circ})$$

Answer 2

A reasonably-accurate diagram really helps.