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Brazilian Mathematical Olympiad

Geometry Difficulty 4.5 AIME Prove it Brazil

$ABCD$ is a convex quadrilateral with $\angle BAC = 30^\circ$ , $\angle CAD = 20^\circ$ , $\angle ABD = 50^\circ$ , $\angle DBC = 30^\circ$ . If the diagonals intersect at $P$ , show that $PC = PD$ .

Solution

We have $\angle ADB = 80^\circ$ , $\angle ACB = 70^\circ$ . Applying the sine rule repeatedly we have $PD = \frac{\sin 20^\circ}{\sin 80^\circ} PA$ , $PC = \frac{\sin 30^\circ}{\sin 70^\circ} PB = \frac{\sin 30^\circ}{\sin 70^\circ} \frac{\sin 30^\circ}{\sin 50^\circ} PA$ . So we have to show that $\sin 30^\circ \sin 30^\circ \sin 80^\circ = \sin 70^\circ \sin 50^\circ \sin 20^\circ$ . Indeed we have $\sin 80^\circ = 2 \sin 40^\circ \cos 40^\circ = 4 \sin 20^\circ \cos 20^\circ \cos 40^\circ = 4 \sin 20^\circ \sin 70^\circ \sin 50^\circ$ and $\sin 30^\circ \sin 30^\circ = \frac{1}{4}$ .

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