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:

from which follows that

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

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

Now if $ABCD$ is cyclic we get

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

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

Multiplication with $AD$ gives:

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

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

Again we use our formulas to get:

From this we get:

Putting this together we get Ptolomy’s inequality:

with equality iff $ABCD$ is cyclic.