Problem
GEO-B2-M02-P024 The Common Chord as a Line of Equal Powers
#24
★★★★★ Level 5 of 5
Two circles intersect at points \(A\) and \(B\). Point \(P\) lies on line \(AB\) outside both circles. A line through \(P\) meets the first circle at \(C,D\), and the second at \(E,F\). Prove that \(PC\cdot PD=PE\cdot PF\).
Since \(P\) lies on the common chord \(AB\), its powers with respect to the two circles are equal.
For the first circle, the power of point \(P\) equals \(PC\cdot PD\). For the second circle, it equals \(PE\cdot PF\). But \(P\) lies on line \(AB\), and \(A\) and \(B\) are common points of the circles. Therefore both powers are also equal to the same product \(PA\cdot PB\). Hence \(PC\cdot PD=PE\cdot PF\).
The final problem connects power of a point with the future radical axis, while staying within this module.