# proof of Ptolemy’s inequality

Looking at the quadrilateral $ABCD$ we construct a point $E$, such that the triangles $ACD$ and $AEB$ are similar ($\angle ABE=\angle CDA$ and $\angle BAE=\angle CAD$).

This means that:

 $\frac{AE}{AC}=\frac{AB}{AD}=\frac{BE}{DC},$

from which follows that

 $BE=\frac{AB\cdot DC}{AD}.$

Also because $\angle EAC=\angle BAD$ and

 $\frac{AD}{AC}=\frac{AB}{AE}$

the triangles $EAC$ and $BAD$ are similar. So we get:

 $EC=\frac{AC\cdot DB}{AD}.$

Now if $ABCD$ is cyclic we get

 $\angle ABE+\angle CBA=\angle ADC+\angle CBA=180^{\circ}.$

This means that the points $C$, $B$ and $E$ are on one line and thus:

 $EC=EB+BC$

Now we can use the formulas we already found to get:

 $\frac{AC\cdot DB}{AD}=\frac{AB\cdot DC}{AD}+BC.$

Multiplication with $AD$ gives:

 $AC\cdot DB=AB\cdot DC+BC\cdot AD.$

Now we look at the case that $ABCD$ is not cyclic. Then

 $\angle ABE+\angle CBA=\angle ADC+\angle CBA\neq 180^{\circ},$

so the points $E$, $B$ and $C$ form a triangle and from the triangle inequality we know:

 $EC

Again we use our formulas to get:

 $\frac{AC\cdot DB}{AD}<\frac{AB\cdot DC}{AD}+BC.$

From this we get:

 $AC\cdot DB

Putting this together we get Ptolomy’s inequality:

 $AC\cdot DB\leq AB\cdot DC+BC\cdot AD,$

with equality iff $ABCD$ is cyclic.

Title proof of Ptolemy’s inequality ProofOfPtolemysInequality 2013-03-22 12:46:09 2013-03-22 12:46:09 mathwizard (128) mathwizard (128) 5 mathwizard (128) Proof msc 51-00