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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 $BC$.

enter image description here

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  • $\begingroup$ Where can we find an image of the problem? $\endgroup$ Commented Apr 22, 2018 at 7:54
  • $\begingroup$ There was no image given, but I think te picture i inserted is correct $\endgroup$
    – SuperMage1
    Commented Apr 22, 2018 at 8:04

2 Answers 2

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A reasonably-accurate diagram really helps.

enter image description here

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

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