Ptolemy's Theorem and Inequality

Ptolemy's Inequality

Ptolemy's Inequality tells us relation between the sides of a quadrilateral and its diagonals. Equality case of this inequality gives a important criterion for a quadrilateral to be cyclic and is also a very important and useful property of cyclic quadrilaterals.

Ptolemy's Inequality says that $$\begin{align*}AB \cdot CD + AD \cdot BC &\ge AC \cdot BD\\\text{(equality occurs} &\Leftrightarrow \square ABCD \text{ is cyclic)}\end{align*}$$

The equality condition is also called as Ptolemy's Theorem: $$AB \cdot CD + AD \cdot BC = AC \cdot BD \Leftrightarrow \square ABCD \, \text{is cyclic.}$$

I'll provide two proofs of Ptolemy's theorem. One uses similar triangles and is slightly artificial, and while other is more natural using inversion.

Proof Using Similar Triangles

We want to use triangle inequality somehow. We do it using a construction which is not a very natural.

We construct point $K$ such that $K$ and $A$ are on same side of $\overline{BC}$ and $\triangle BCK \sim \triangle BDA$

Thus we have $\frac{BC}{BD} = \frac{BK}{BA} \, \, \Rightarrow \frac{BC}{BK} = \frac{BD}{BA}$ Also, $\angle CBD = \angle KBA$ Thus, by $SAS$ test of similarity, $\triangle CBD \sim \triangle KBA$.

So we have $\angle CKB = \angle A$ and $\angle AKB = \angle C$.

$CK = \dfrac{BC \cdot AD}{BD}$ and $AK = \dfrac{AB \cdot CD}{BD}$

Triangle inequality gives $CK + AK \ge AC \Rightarrow \dfrac{BC \cdot AD}{BD} + \dfrac{AB \cdot CD}{BD} \ge AC \Rightarrow BC \cdot AD + AB \cdot CD \ge AC \cdot BD$

Also note that equality occurs only when $K$ lies on $\overline{AC}$. That is when $\angle CKB + \angle BKA = 180^{\circ} \Leftrightarrow \angle A + \angle C = 180^{\circ} \Leftrightarrow \square ABCD \text{is cyclic.}$

So we have the required result!!

Remark: Here we used $SAS$ test of similarity which is not used very commonly, and is bit tricky to use compared to the other tests for similarity of triangles.

Proof Using Inversion

We invert in one of the vertices say in $A$ with suitable radius say $r$. So that we get following figure after inversion.

Where $B \rightarrow X, \, C\rightarrow Y, \, D \rightarrow Z$ after inversion.

Hence once again by triangle inequality we have $XY + YZ \ge XZ$

$\Rightarrow \dfrac{BC \cdot r^2}{AB \cdot AC} + \dfrac{CD \cdot r^2}{AC \cdot AD} \ge \dfrac{BD \cdot r^2}{AB \cdot AD} \Rightarrow BC \cdot AD + CD \cdot AB \ge AC \cdot BD$

Equality when the three points are collinear, that is when $\square ABCD$ is cyclic.

There is a generalization of Ptolemy's theorem called Casey's Theorem.

Some Problems

Please add more problems with disguised Ptolemy applications here. Two above are very straightforward.