Problem
GEO-B2-M02-P006 A Product Creates a Circle
#6
★★★☆☆ Level 3 of 5
Points \(A,B\) lie on one ray starting at \(P\), and points \(C,D\) lie on another ray. It is known that \(PA\cdot PB=PC\cdot PD\). Prove that points \(A,B,C,D\) lie on one circle.
Draw the circle through \(A,B,C\) and look where it meets ray \(PC\) for the second time.
Let the circle through \(A,B,C\) meet ray \(PC\) for the second time at \(D'\). By the secant theorem, \(PA\cdot PB=PC\cdot PD'\). By the condition, \(PA\cdot PB=PC\cdot PD\), hence \(PD'=PD\). Since \(D\) and \(D'\) lie on the same ray from \(P\), they coincide. Therefore \(A,B,C,D\) are cyclic.
An important reverse technique: prove a circle by a product, not by angles.