Problem
GEO-B2-M02-P016 Intersection of Extended Sides
#16
★★★★☆ Level 4 of 5
In cyclic quadrilateral \(ABCD\), lines \(AB\) and \(CD\) meet at point \(P\) outside the circle. Prove that \(PA\cdot PB=PC\cdot PD\).
From point \(P\), two secants to one circle are drawn.
Line \(PAB\) meets the circle at \(A\) and \(B\), while line \(PCD\) meets it at \(C\) and \(D\). Therefore, by the two-secant theorem, \(PA\cdot PB=PC\cdot PD\).
The student must see the external point and two secants, even when they are sides of the quadrilateral.