Problem
GEO-B2-M02-P005 Prove the Secant Formula
From point \(P\) outside a circle, secants \(PAB\) and \(PCD\) are drawn, where \(A\) and \(C\) are the nearer points. Prove that \(PA\cdot PB=PC\cdot PD\).
Join \(A\) to \(C\) and \(B\) to \(D\), then compare \(\triangle PAC\) and \(\triangle PDB\).
Join \(A\) to \(C\), and \(B\) to \(D\). Consider triangles \(PAC\) and \(PDB\). The angle at \(P\) is common, because \(P,A,B\) and \(P,C,D\) are collinear. Also, \(\angle PAC=\angle BAC\), \(\angle PDB=\angle CDB\), and angles \(\angle BAC\) and \(\angle CDB\) are equal as inscribed angles standing on chord \(BC\). Therefore \(\triangle PAC\sim\triangle PDB\). From similarity, \(\frac{PA}{PD}=\frac{PC}{PB}\), hence \(PA\cdot PB=PC\cdot PD\).
The student should understand where the formula comes from, not only substitute numbers.