proof of Ptolemy’s theorem

Let ABCD be a cyclic quadrialteral. We will prove that


Find a point E on BD such that BCA=ECD. Since BAC=BDC for opening the same arc, we have triangle similarityMathworldPlanetmath ABCDEC and so


which implies ACED=ABCD.

Also notice that ADCBEC since have two pairs of equal angles. The similarity implies


which implies ACBE=BCDA.

So we finally have ACBD=AC(BE+ED)=ABCD+BCDA.

Title proof of Ptolemy’s theorem
Canonical name ProofOfPtolemysTheorem
Date of creation 2013-03-22 12:38:31
Last modified on 2013-03-22 12:38:31
Owner drini (3)
Last modified by drini (3)
Numerical id 11
Author drini (3)
Entry type Proof
Classification msc 51-00
Related topic PtolemysTheorem
Related topic CyclicQuadrilateral