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Singapore Mathematical Olympiad (SMO)

Geometry Difficulty 4.8 AIME Prove it Singapore

The quadrilateral $ABCD$ is inscribed in a circle which has diameter $BD$ . Points $A'$ and $B'$ are symmetric to $A$ and $B$ with respect to the line $BD$ and $AC$ respectively. If the lines $A'C$ , $BD$ intersect at $P$ and $AC$ , $B'D$ intersect at $Q$ , prove that $PQ$ is perpendicular to $AC$ .

Solution

Let $AC$ intersect $BD$ at $R$ . Then $\angle BAR = \angle BAC = \angle BA'P = \angle BAP$ . That is $AB$ bisects $\angle PAR$ . As $\angle BAD = 90^\circ$ , we also have $AD$ is the external bisector of $\angle PAR$ . By the angle bisector theorem, we have
$\frac{BR}{BP} = \frac{DR}{DP} = \frac{AR}{AP} \qquad (1)$
As $B$ and $B'$ are symmetric with respect to the line $AC$ , we have $\angle BQR = \angle B'QR = \angle DQR$ . Thus $QR$ bisects $\angle BQD$ . By the angle bisector theorem and (1), we have
$\frac{QD}{QB} = \frac{RD}{RB} = \frac{PD}{PB}.$
Thus $QP$ is the external bisector of $\angle BQD$ . Hence $\angle RQP = 90^\circ$ .

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