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Property of Perpendicular Chords

Property of Perpendicular Chords

IP 3

\(AC\) and \(BD\) are chords of a circle with radius \(r\). \(AC\), \(BD\) intersect at \(P\) and \(AC \perp BD\), show that \(AP^2 + BP^2 + CP^2 + DP^2 = 4r^2\).

A circle with center \(O\) and two perpendicular chords \(AC\) and \(BD\) intersecting at point \(P\).

Solution:

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Answer:\(AP^2+BP^2+CP^2+DP^2=4r^2\), as proved using perpendicular chord bisectors and Pythagoras.

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