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 Theorem

If 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:

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

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. \(_\square \)

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

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

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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,

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

Submit your answer

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?

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\(ABCD\) 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.

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A cyclic quadrilateral \(ABCD\) is constructed within a circle such that \(AB = 3, BC = 6,\) and \(\triangle ACD\) is equilateral, as shown to the right.

If \(E\) is the intersection point of both diagonals of \(ABCD\), what is the length of \(ED,\) the blue line segment in the diagram?

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.

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. \(_\square\)