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: $$EC<EB+BC.$$ Again we use our formulas to get: $$\frac{AC\cdot DB}{AD}<\frac{AB\cdot DC}{AD}+BC.$$ From this we get: $$AC\cdot DB<AB\cdot DC+BC\cdot AD.$$ Putting this together we get Ptolomy's inequality: $$AC\cdot DB\leq AB\cdot DC+BC\cdot AD,$$ with equality iff $ABCD$ is cyclic.