Problem
GEO-B2-M02-P020 Prove a Circle From Two Products
#20
★★★★☆ Level 4 of 5
Lines \(l_1\) and \(l_2\) meet at point \(P\). Points \(A,B\) are chosen on \(l_1\), and points \(C,D\) on \(l_2\), with \(P\) not between the points of each pair. If \(PA\cdot PB=PC\cdot PD\), prove that \(A,B,C,D\) lie on one circle.
This is the converse power criterion, but the position of the points must be used carefully.
Draw the circle through \(A,B,C\). Let it meet line \(l_2\) for the second time at \(D'\), located on the same ray from \(P\) as \(D\). By the secant theorem, \(PA\cdot PB=PC\cdot PD'\). Comparing with the condition, we get \(PD'=PD\), hence \(D'=D\). Therefore \(A,B,C,D\) lie on one circle.
A more mature formulation of the converse criterion, without a triangular hint.