Problem
GEO-B2-M10-P007 Diagonals of a Cyclic Quadrilateral
#7
★★★☆☆ Level 3 of 5
In a convex cyclic quadrilateral \(ABCD\), diagonals \(AC\) and \(BD\) meet at \(P\). Prove that \(PA\cdot PC=PB\cdot PD\).
Point \(P\) has the same power with respect to the circle along two secants.
Lines \(PAC\) and \(PBD\) are two secants of the same circle. By power of a point, \(PA\cdot PC=PB\cdot PD\).
This problem reinforces that power of a point also works for a point inside the circle.