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 7 of 7: Conclude the collinearity
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RDRA=R′DR′A⟹R=R′⟹B,E,R are collinear.\frac{RD}{RA}=\frac{R'D}{R'A}\Longrightarrow R=R'\Longrightarrow B,E,R\text{ are collinear}.
Detailed analysis

On the fixed line AD, the point determined by the directed ratio to A and D is unique. Hence R=R', which is exactly the statement that B,E,R are collinear.