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 3 of 7: Apply Ptolemy to ABCD
Detailed analysis
The ratio from the previous step gives AB·DC=BC·AD. Ptolemy's theorem says AB·DC+BC·AD=AC·BD, so each equal term is half AC·BD. Rearranging gives DB/AB=2DC/CA.