A cyclic quadrilateral \(ABCD\) has \(AB=BC\). Let the intersection of \(AC\) and \(BD\) be \(P\). In addition, let the line through \(D\) tangent to the circumcircle of \(ABCD\) intersect the extension of \(AC\) at \(Q\).

Prove that \(PQ=DQ\).

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## Comments

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TopNewestThis is the quickest proof that I know of. Draw the tangent through \(B\). Since \(AB=BC\), then the tangent is parallel to \(QC\). But the tangent through \(B\) makes an angle with \(BD\) the same as the tangent through \(D\) makes with \(BD\). Therefore, since this is the same as the angle \(QC\) makes with \(BD\), triangle \(DQP\) is isosceles and \(PQ=DQ\).

An annotated picture is worth a thousand words more than an un-annotated picture.

Lui's Cyclic

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Can you add this to the Brilliant Wiki, under a relevant skill in Circles? Thanks!

@Daniel Liu

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This is just angle chasing. We have that \(\angle QDP = \angle DCB\) and \[\angle QPD = \angle CPB = 180^{\circ} - \angle PCB - \angle PBC = 180^{\circ} - \angle PAB - \angle DAP \\ = 180^{\circ} - \angle DAB = \angle DCB = \angle QDP.\] It hence follows that \(QD=QP.\) \(\blacksquare\)

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