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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 $$PQ=DQ$$.

Note by Daniel Liu
2 years, 10 months ago

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This 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

· 2 years, 10 months ago

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

@Daniel Liu Staff · 2 years, 9 months ago

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 $$QD=QP.$$ $$\blacksquare$$ · 2 years, 10 months ago