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 $\mathrm{\angle }BCA=\mathrm{\angle }ECD$. Since $\mathrm{\angle }BAC=\mathrm{\angle }BDC$ for opening the same arc, we have triangle similarity $\mathrm{△}ABC\sim \mathrm{△}DEC$ and so

$$\frac{AB}{DE}=\frac{CA}{CD}$$

which implies $AC\cdot ED=AB\cdot CD$.

Also notice that $\mathrm{△}ADC\sim \mathrm{△}BEC$ 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
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