Intersecting Chords Theorem - a Visual Proof
Given a point \(P\) in the interior of a circle, pass two lines through \(P\) that intersect the circle in points \(A\) and \(D\) and, respectively, \(B\) and \(C\). Then \(AP\cdot DP = BP\cdot CP\).
If \(a=AP\), \(b=BP\), \(c=CP\), and \(d=DP\), then the Intersecting Chors Theorem is expressed as
\(a\cdot d = b\cdot c\).
(The applet below illustrates the proof. Points \(A\), \(B\), \(C\), \(D\), \(O\), \(R\) are draggable. Point \(O\) is the center of the given circle, \(R\) - a point on the circle.)
This proof is a slightly modified version of the one suggested by Hubert Shutrick.
Apply homothety with center \(P\) and the coefficient \(b\) to \(\Delta APC\) and another with the coefficient \(a\) to \(\Delta BPD\).
Observe that, as two inscribed angles subtended by the same arc, \(\angle CAD =\angle CBD\), and, similarly, \(\angle ACB =\angle ADB\). Since homotheties preserves angles, we also have \(\angle C'A'D' =\angle C'B'D'\) and \(\angle A'C'B' =\angle A'D'B'\). By the construction, \(A'P = ba = ab = B'P\), making triangles \(A'PC'\) and \(B'PD'\) equal, so that \(C'P=D'P\) but \(C'P=bc\) whereas \(D'P=ad\). Therefore, \(bc=ad\).
Copyright © 1996-2018 Alexander Bogomolny