View source for Ptolemy's theorem

'''Ptolemy's theorem''' gives a relationship between the side lengths and the diagonals of a [[cyclic quadrilateral]]; it is the [[equality condition | equality case]] of [[Ptolemy's Inequality]]. Ptolemy's theorem frequently shows up as an intermediate step in problems involving inscribed figures.

== Statement ==

Given a [[cyclic quadrilateral]] <math>ABCD</math> with side lengths <math>{a},{b},{c},{d}</math> and [[diagonal]]s <math>{e},{f}</math>:

<cmath>ac+bd=ef.</cmath>

== Proof 1 ==

Given cyclic quadrilateral <math>ABCD,</math> extend <math>CD</math> to <math>P</math> such that <math>\angle BAC=\angle DAP.</math> 

Since quadrilateral <math>ABCD</math> is cyclic, <math>m\angle ABC+m\angle ADC=180^\circ .</math> However, <math>\angle ADP</math> is also supplementary to <math>\angle ADC,</math> so <math>\angle ADP=\angle ABC</math>. Hence, <math>\triangle ABC \sim \triangle ADP</math> by AA similarity and <math>\frac{AB}{AD}=\frac{BC}{DP}\implies DP=\frac{(AD)(BC)}{(AB)}.</math>

Now, note that <math>\angle ABD=\angle ACD </math> (subtend the same arc) and <math>\angle BAC+\angle CAD=\angle DAP+\angle CAD \implies \angle BAD=\angle CAP,</math> so <math>\triangle BAD\sim \triangle CAP.</math> This yields <math>\frac{AB}{AC}=\frac{BD}{CP}\implies CP=\frac{(AC)(BD)}{(AB)}.</math>

However, <math>CP= CD+DP.</math> Substituting in our expressions for  <math>CP</math> and  <math>DP,</math>  <math> \frac{(AC)(BD)}{(AB)}=CD+\frac{(AD)(BC)}{(AB)}.</math> Multiplying by <math>AB</math> yields  <math>(AC)(BD)=(AB)(CD)+(AD)(BC)</math>.


== Proof 2 (inversion) == 

We provide a proof for the general case of Ptolemy's theorem, Ptolemy's Inequality. 

Let <math>A,B,C,D</math> be four points in the Euclidean plane. Taking an inversion centered at <math>D</math> (the point doesn't matter, it can be any of the four) with radius <math>r</math>, we have that <math>A^*B^*+B^*C^*\geq A^*C^*</math> by the Triangle Inequality, with equality holding when <math>A^*, B^*, C^*</math> are collinear, i.e. when <math>A,B,C</math> lie on a circle containing <math>D.</math> Additionally, by the Inversion Distance Formula, we may express the inequality as the following: 

<cmath>\frac{r^2}{AD\cdot BD}\cdot AB + \frac{r^2}{BD\cdot CD}\cdot BC \geq \frac{r^2}{AD\cdot CD}\cdot AC.</cmath>

Dividing by <math>r^2</math> and multiplying everything by <math>AD\cdot BD \cdot CD,</math> we get <math>AB\cdot CD + BC\cdot AD \geq AC\cdot BD,</math> and thus the desired. <math>_\blacksquare</math>

== Problems ==
===2023 AIME I Problem 5===
Square <math>ABCD</math> is inscribed in a circle. Point <math>P</math> is on this circle such that <math>AP \cdot CP = 56</math>, and <math>BP \cdot DP = 90</math>. What is the area of the square?

([[2023 AIME I Problems/Problem 5|Source]])

===2004 AMC 10B Problem 24===
In triangle <math>ABC</math> we have <math>AB=7</math>, <math>AC=8</math>, <math>BC=9</math>. Point <math>D</math> is on the circumscribed circle of the triangle so that <math>AD</math> bisects angle <math>BAC</math>. What is the value of <math>AD/CD</math>?

<math>\text{(A) } \dfrac{9}{8} \quad \text{(B) } \dfrac{5}{3} \quad \text{(C) } 2 \quad \text{(D) } \dfrac{17}{7} \quad \text{(E) } \dfrac{5}{2}</math>

([[2004 AMC 10B Problems/Problem 24|Source]])

=== Equilateral Triangle Identity ===
Let <math>\triangle ABC</math> be an equilateral triangle. Let <math>P</math> be a point on minor arc <math>AB</math> of its circumcircle. Prove that <math>PC=PA+PB</math>.

Solution: Draw <math>PA</math>, <math>PB</math>, <math>PC</math>. By Ptolemy's theorem applied to quadrilateral <math>APBC</math>, we know that <math>PC\cdot AB=PA\cdot BC+PB\cdot AC</math>. Since <math>AB=BC=CA=s</math>, we divide both sides of the last equation by <math>s</math> to get the result: <math>PC=PA+PB</math>.

=== Regular Heptagon Identity ===
In a regular heptagon <math> ABCDEFG </math>, prove that: <math> \frac{1}{AB}=\frac{1}{AC}+\frac{1}{AE} </math>.

Solution: Let <math> ABCDEFG </math> be the regular heptagon. Consider the quadrilateral <math> ABCE </math>. If <math> a </math>, <math> b </math>, and <math> c </math> represent the lengths of the side, the short diagonal, and the long diagonal respectively, then the lengths of the sides of <math> ABCE </math> are <math> a </math>, <math> a </math>, <math> b </math> and <math> c </math>; the diagonals of <math> ABCE </math> are <math> b </math> and <math> c </math>, respectively.

Now, Ptolemy's theorem states that <math> ab + ac = bc </math>, which is equivalent to <math> \frac{1}{a}=\frac{1}{b}+\frac{1}{c} </math> upon division by <math> abc </math>.

=== 1991 AIME Problems/Problem 14 ===
A hexagon is inscribed in a circle. Five of the sides have length <math>81</math> and the sixth, denoted by <math>\overline{AB}</math>, has length <math>31</math>. Find the sum of the lengths of the three diagonals that can be drawn from <math>A</math>.

([[1991_AIME_Problems/Problem_14|Source]])

=== Cyclic Hexagon ===
A hexagon with sides of lengths 2, 2, 7, 7, 11, and 11 is inscribed in a circle.  Find the diameter of the circle.

Solution: Consider half of the circle, with the quadrilateral <math>ABCD</math>, <math>AD</math> being the diameter. <math>AB = 2</math>, <math>BC = 7</math>, and <math>CD = 11</math>. Construct diagonals <math>AC</math> and <math>BD</math>. Notice that these diagonals form right triangles. You get the following system of equations:

<math>(AC)(BD) = 7(AD) + 22</math> (Ptolemy's theorem)

<math>\text(AC)^2 = (AD)^2 - 121</math>

<math>(BD)^2 = (AD)^2 - 4</math>

Solving gives <math>AD = 14</math>

== See also ==
* [[Geometry]]
* [[Cyclic quadrilateral]]

[[Category:Geometry]]
[[Category:Theorems]]