ISL 2021 G1. Let $ABCD$ be a parallelogram such that $AC=BC$ . A point $P$ is chosen on the extension of segment $AB$ beyond $B$ . Define $Q=(ACD)\cap PD$ and $R=(APQ)\cap PC$ . Prove that $CD$ , $AQ$ , and $BR$ are concurrent.

Solution. Define $E=AQ\cap CD$ . Use directed angles modulo $360^\circ$ .

Claim 1. $C$ , $Q$ , $R$ , $E$ are concyclic.

Proof. We have $\begin{align*}\measuredangle CEQ&=180^\circ-\measuredangle EAD-\measuredangle ADE\\&=180^\circ-(\measuredangle QAC+\measuredangle CAD)-(\measuredangle ADQ+\measuredangle QDC)\\&=180^\circ-\measuredangle QAC-\measuredangle QDC-\measuredangle CAD-\measuredangle BAQ\\&=180^\circ-2\measuredangle QAC-(180^\circ-\measuredangle ADC-\measuredangle DCA)-\measuredangle BAQ\\&=\measuredangle ADC+\measuredangle DCA-2\measuredangle QAC-\measuredangle BAQ\\&=2\measuredangle ADC-2\measuredangle QAC-\measuredangle BAQ\\&=2\measuredangle ADQ+2\measuredangle QDC-2\measuredangle QAC-\measuredangle BAQ\\&=2\measuredangle ADQ-\measuredangle BAQ\\&=\measuredangle BAQ\\&=\measuredangle CRQ.\end{align*}$
Claim 2. $B$ , $R$ , $E$ are collinear.

Proof. We have $\begin{align*}\measuredangle QRB&=180^\circ-\measuredangle CRQ-\measuredangle ERC\\&=180^\circ-\measuredangle PAQ-\measuredangle EQC\\&=180^\circ-\measuredangle PAQ-\measuredangle ADC\\&=180^\circ-\measuredangle ACQ-\measuredangle DCA\\&=\measuredangle QCE\\&=180^\circ-\measuredangle ERQ.\end{align*}$
Now, Claim 2 implies the result.

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