# proof of Ptolemy’s theorem

Let $ABCD$ be a cyclic quadrialteral. We will prove that

 $AC\cdot BD=AB\cdot CD+BC\cdot DA.$

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 $AC\cdot BD=AC(BE+ED)=AB\cdot CD+BC\cdot DA$.

Title proof of Ptolemy’s theorem ProofOfPtolemysTheorem 2013-03-22 12:38:31 2013-03-22 12:38:31 drini (3) drini (3) 11 drini (3) Proof msc 51-00 PtolemysTheorem CyclicQuadrilateral