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Singapore Mathematical Olympiad (SMO)

Geometry Difficulty 6.3 National Olympiad Prove it Singapore

In triangle $ABC$ , $\angle B = 90^\circ$ , $AB > BC$ , and $P$ is the point such that $BP = BC$ and $\angle APB = 90^\circ$ , where $P$ and $C$ lie on the same side of $AB$ . Let $Q$ be the point on $AB$ such that $AP = AQ$ , and let $M$ be the midpoint of $QC$ . Prove that the line through $M$ parallel to $AP$ passes through the midpoint of $AB$ .

Solution

Let $\angle BPC = \angle BCP = \alpha$ , $\angle CBP = \beta$ . As $\angle QAP = 90^\circ - \angle PBA = \angle CBP = \beta$ . Therefore $\angle APQ = \angle AQP = \alpha$ , since $AP = AQ$ .

Also $\angle CPQ = \alpha + \angle BPQ = \angle APB = 90^\circ$ . Therefore the points $B$ , $C$ , $P$ , $Q$ all lie on a circle with centre $M$ which is the midpoint of $QC$ . Thus $\angle QBM = \angle BQC = \angle BPC = \alpha$ .

Since $MN \parallel PA$ , $\angle BNM = \angle QAP = \beta$ . Therefore $\angle BMN = \alpha$ and it follows that $BN = MN$ .

Since $AP = AQ$ and $MP = MQ$ , $AM$ bisects $\angle QAP = \beta$ . Since $\angle BNM = \beta$ , $\angle AMN = \angle MAN = \beta/2$ . Therefore $MN = AN$ . Thus $N$ is the midpoint of $AB$ .

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