MathLabs

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 6 of 7: Compute the ratio at R'
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R′DR′A=ED⋅DBEA⋅AB=EDEA⋅2DCCA=(2EDAE)2=RDRA.\frac{R'D}{R'A}=\frac{ED\cdot DB}{EA\cdot AB}=\frac{ED}{EA}\cdot\frac{2DC}{CA}=\left(\frac{2ED}{AE}\right)^2=\frac{RD}{RA}.
Detailed analysis

Similar triangles ABR' and EDR' give R'D/R'B=ED/AB; similar triangles DBR' and EAR' give R'A/R'B=EA/DB. Dividing and using the two Ptolemy ratios produces the same value as RD/RA.