proof of Ptolemy's theoremProofOfPtolemysTheorem2002-05-15 23:13:112002-05-16 04:15:47ProofPtolemy's theorem\usepackage{graphicx}
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\newcommand{\mathbb}[1]{\mathbbmss{#1}}Let $ABCD$ be a cyclic quadrialteral. We will prove that
$$AC\cdot BD=AB\cdot CD + BC\cdot DA.$$
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Find a point $E$ on $BD$ such that $\angle BCA=\angle ECD$. Since $\angle BAC=\angle BDC$ for opening the same arc, we have triangle similarity
$\triangle ABC\sim \triangle DEC$ and so
$$\frac{AB}{DE}=\frac{CA}{CD}$$
which implies $AC\cdot ED = AB\cdot CD$.
Also notice that $\triangle ADC \sim \triangle BEC$ since have two pairs of equal angles. The similarity implies
$$\frac{AC}{BC}=\frac{AD}{BE}$$
which implies $AC\cdot BE = BC\cdot DA$.
So we finally have $AC\cdot BD=AC(BE+ED)=AB\cdot CD+BC\cdot DA$.