South-Afrika, 2011

Geometry Difficulty 6.0 National Olympiad Prove it South Africa

Let $ABC$ be a triangle with orthocentre $H$ and let $P$ be a point on its circumcircle. The line through $A$ parallel to $BP$ meets $CH$ at $Q$ and the line through $A$ parallel to $CP$ meets $BH$ at $R$ . Prove that $QR$ is parallel to $AP$ .

Solution

Let $T$ be the intersection of $BH$ and $PC$ and $F$ be the intersection of $AB$ and $CH$ . Then
$\begin{aligned} \angle HQA &= 90^\circ - \angle BAQ \\ &= 90^\circ - \angle ABP \quad (\text{since } PB \parallel AQ) \\ &= 90^\circ - \angle ACP \\ &= \angle RTC \\ &= \angle ART \quad (\text{since } PC \parallel AR) \end{aligned}$

which shows that AHQR is a cyclic quadrilateral. Hence
$\angle ARQ = \angle AHF = \angle ABC = \angle APC.$

Now, since $AR \parallel PC$ , it follows that $AP \parallel QR$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.