# Ptolemy's Theorem

**Ptolemy's theorem** states the relationship between the diagonals and the sides of a cyclic quadrilateral. It is a powerful tool to apply to problems about inscribed quadrilaterals.

Ptolemy's TheoremIf a quadrilateral is inscribable in a circle, then the product of the measures of its diagonals is equal to the sum of the products of the measures of the pairs of the opposite sides:

\[AC\cdot BD = AB\cdot CD + AD\cdot BC.\] \(\hspace{1.5cm}\)

#### Contents

## Proof

Let's prove this theorem.

Let \(ABCD\) be a random quadrilateral inscribed in a circle.

The proposition will be proved if \(AC\cdot BD = AB\cdot CD + AD\cdot BC.\)

It's easy to see in the inscribed angles that \(\angle ABD = \angle ACD, \angle BDA= \angle BCA,\) and \(\angle BAC = \angle BDC. \)

Let \(E\) be a point on \(AC\) such that \( \angle EBC = \angle ABD = \angle ACD,\) then since \( \angle EBC = \angle ABD \) and \(\angle BCA= \angle BDA,\)

\[\triangle EBC \approx \triangle ABD \Leftrightarrow \dfrac{CB}{DB} = \dfrac{CE}{AD} \Leftrightarrow AD\cdot CB = DB\cdot CE. \qquad (1)\]

Note that \(\angle ABD = \angle EBC \Leftrightarrow \angle ABD + \angle KBE = \angle EBC + \angle KBE \Rightarrow \angle ABE = \angle CBK.\) Then since \(\angle ABE= \angle CBK\) and \(\angle CAB= \angle CDB,\)

\[\triangle ABE \approx \triangle BDC \Leftrightarrow \dfrac{AB}{DB} = \dfrac{AE}{CD} \Leftrightarrow CD\cdot AB = DB\cdot AE. \qquad (2)\]

Therefore, from \((1)\) and \((2),\) we have

\[\begin{align} AB \cdot CD + AD\cdot BC & = CE\cdot DB + AE\cdot DB \\ & = (CE+AE)DB \\ & = CA\cdot DB. \end{align}\]

Hence proved. \(_\square \)

## Applications

We can prove the Pythagorean theorem using Ptolemy's theorem:

Prove that in any right-angled triangle \(\triangle ABC\) where \(\angle A = 90^\circ,\) \(AB^2 + AC^2 = BC^2.\)

Let \(ABDC\) be a random rectangle inscribed in a circle.\[\]

\[\]Applying Ptolemy's theorem in the rectangle, we get

\[AD\cdot BC = AB\cdot DC + AC\cdot DB.\]

But \(AD= BC, AB = DC, AC = DB\) since \(ABDC\) is a rectangle. Therefore,

\[BC^2 = AB^2 + AC^2. \ _\square\]

Once upon a time, Ptolemy let his pupil draw an equilateral triangle \(ABC\) inscribed in a circle before the great mathematician depicted point \(D\) and joined the red lines with other vertices, as shown below.

**Ptolemy**: Dost thou see that all the red lines have the lengths in whole integers?

**Pupil**: Indeed, master! Such an extraordinary point!

**Ptolemy**: Now if the equilateral triangle has a side length of 13, what is the sum of the three red lengths combined?

\(ABCD\) is a cyclic quadrilateral with \(\displaystyle \overline{AB}=11\) and \(\displaystyle \overline{CD}=19\). \(P\) and \(Q\) are points on \(\overline{AB}\) and \( \overline{CD}\), respectively, such that \(\displaystyle \overline{AP}=6\), \(\displaystyle \overline{DQ}=7\), and \(\displaystyle \overline{PQ}=27.\) Determine the length of the line segment formed when \(\displaystyle \overline{PQ}\) is extended from both sides until it reaches the circle.

**Note**: The image is not drawn to scale.

## Converse

In a quadrilateral, if the product of its diagonals is equal to the sum of the products of the pairs of the opposite sides, then the quadrilateral is

inscribable.

## Ptolemy's Inequality

Given a quadrilateral \(ABCD\),

\[AC\cdot BD \leq AB\cdot CD + AD\cdot BC,\]

where equality occurs if and only if \(ABCD\) is inscribable. \(_\square\)

Let \(I\) be a point inside quadrilateral \(ABCD\) such that \(\angle ABD = \angle IBC\) and \(\angle ADB = \angle ICB\).

Triangle \(ABD\) is similar to triangle \(IBC\), so \(\frac{AB}{IB}=\frac{BD}{BC}=\frac{AD}{IC} \implies AD \cdot BC = BD \cdot IC\) and \(\frac{AB}{BD}=\frac{IB}{BC}\).

This gives us another pair of similar triangles: \(ABI\) and \(DBC\) \(\implies \frac{AI}{DC}=\frac{AB}{BD} \implies AB \cdot CD = AI \cdot BD\).

Adding the two equations together,

\[AB\cdot CD + AD\cdot BC = BD \cdot (IA + IC) \geq BD \cdot AC.\]

The equality occurs when \(I\) lies on \(AC\), which means \(ABCD\) is inscribable. \(_\square\)

Another proof requires a basic understanding of properties of inversions, especially those relevant to distance.

Consider a circle of radius 1 centred at \(A\). Let \(B', C',\) and \(D'\) be the resultant of inverting points \(B, C,\) and \(D\) about this circle, respectively. Using the distance properties of inversion, we have

\[\begin{align} AB &= \frac{1}{AB'}\\ CD &= \frac{C'D'}{AC' \cdot AD'}\\ AD &= \frac{1}{AD'}\\ BC &= \frac{B'C'}{AB' \cdot AC'}\\ AC &= \frac{1}{AC'}\\ BD &= \frac{B'D'}{AB' \cdot AD'}. \end{align}\]

Thus, Ptolemy's inequality becomes

\[\begin{align} AB \cdot CD + AD \cdot BC &\geq BD \cdot AC\\ \frac{1}{AB'} \cdot \frac{C'D'}{AC' \cdot AD'} + \frac{1}{AD'} \cdot \frac{B'C'}{AB' \cdot AC'} &\geq \frac{1}{AC'} \cdot \frac{B'D'}{AB' \cdot AD'}\\\\ C'D' + B'C' &\geq B'D', \end{align}\]

which is true by triangle inequality. Therefore, Ptolemy's inequality is true. Thus proven. \(_\square\)

**Cite as:**Ptolemy's Theorem.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/ptolemys-theorem/