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 4 of 7: Use Ptolemy again
Detailed analysis
Applying the same tangent-chord and cyclic-angle similarities to the quadrilateral containing A,C,D,E gives CA·ED=CE·AD. Ptolemy then identifies this common product with half AE·DC, yielding the displayed ratio.