Problem 5

Let ABCD be a quadrilateral inscribed in a circle omega, and let P be a point on the extension of AC such that PB and PD are tangent to omega. The tangent at C intersects PD at Q and the line AD at R. Let E be the second point of intersection of AQ and omega. Prove that B, E, R are collinear.

Step 5 of 7: Compute the ratio at R

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\triangle RDC\sim\triangle RCA\Longrightarrow\frac{RD}{RA}=\left(\frac{DC}{CA}\right)^2=\left(\frac{2ED}{AE}\right)^2.

Detailed analysis

The cyclic angle relations make RDC similar to RCA, so RD/RC=DC/CA=RC/RA. Multiplying the first and third ratios gives RD/RA=(DC/CA)^2; substitute the preceding relation.