Ptolemy's Theorem

Ptolemy's theorem states the relationship between the diagonals and the sides of the 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:

\[AC\cdot BD = AB\cdot CD + AD\cdot BC.\ _\square\]

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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 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.\[\]

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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,

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

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

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If it is true in a quadrilateral that 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. \(_ \square\)

Given a quadrilateral \(ABCD\),

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

where equality occurs iff \(ABCD\) 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, \(\dfrac{AB}{IB}=\dfrac{BD}{BC}=\dfrac{AD}{IC} \Rightarrow AD \cdot BC = BD \cdot IC\) and \(\dfrac{AB}{BD}=\dfrac{IB}{BC}\). This gives us another pair of similar triangles: \(ABI\) and \(DBC\) \(\Rightarrow \dfrac{AI}{DC}=\dfrac{AB}{BD} \Rightarrow AB \cdot CD = AI \cdot BD\).

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.