Difference between revisions of "Cross-Segment Identity"

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Latest revision as of 01:35, 2 January 2024

Motivation

After repetitive usage of this in AMC/AIME Euclidean Geometry problems, I (mathboy282) only found it fit to make it an identity.

Identity

Let there be a cyclic quadrilateral $ABCD$ with $E$ as the intersection of the diagonals $AC$ and $BD.$

Let $AE=a, BE=b, CE=c,$ and $DE=d.$ Then, we must have: \[ab+bc+cd+da=AB \cdot CD + BC \cdot DA.\]

Proof

This comes as a direct result of Ptolemy's theorem. \[ab+bc+cd+da=b(a+c)+d(c+a)=(b+d)(a+c)=AB \cdot CD + BC \cdot DA.\]


CrossSegment-Identity.png