proof of Ptolemy’s inequality

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

This means that:

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

from which follows that

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

Also because $\mathrm{\angle }EAC=\mathrm{\angle }BAD$ 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

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

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

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

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:

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

with equality iff $ABCD$ is cyclic.