Ptolemy's Theorem

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


Given a cyclic quadrilateral $ABCD$ with side lengths ${a},{b},{c},{d}$ and diagonals ${e},{f}$:



Given cyclic quadrilateral $ABCD,$ extend $CD$ to $P$ such that $\angle BAD=\angle CAP.$

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

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

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


2004 AMC 10B Problem 24

In triangle $ABC$ we have $AB=7$, $AC=8$, $BC=9$. Point $D$ is on the circumscribed circle of the triangle so that $AD$ bisects angle $BAC$. What is the value of $AD/CD$?

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

Solution: Set $\overline{BD}$'s length as $x$. $CD$'s length must also be $x$ since $\angle BAD$ and $\angle DAC$ intercept arcs of equal length(because $\angle BAD =\angle DAC$). Using Ptolemy's Theorem, $7x+8x=9(AD)$. The ratio is $\boxed{\frac{5}{3}}\implies(B)$

Equilateral Triangle Identity

Let $\triangle ABC$ be an equilateral triangle. Let $P$ be a point on minor arc $AB$ of its circumcircle. Prove that $PC=PA+PB$.

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

Regular Heptagon Identity

In a regular heptagon $ABCDEFG$, prove that: $\frac{1}{AB}=\frac{1}{AC}+\frac{1}{AD}$.

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

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

1991 AIME Problems/Problem 14

A hexagon is inscribed in a circle. Five of the sides have length $81$ and the sixth, denoted by $\overline{AB}$, has length $31$. Find the sum of the lengths of the three diagonals that can be drawn from $A$.


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 $ABCD$, $AD$ being the diameter. $AB = 2$, $BC = 7$, and $CD = 11$. Construct diagonals $AC$ and $BD$. Notice that these diagonals form right triangles. You get the following system of equations:

$(AC)(BD) = 7(AD) + 22$ (Ptolemy's Theorem)

$\text(AC)^2 = (AD)^2 - 121$

$(BD)^2 = (AD)^2 - 4$

Solving gives $AD = 14$

See also

Invalid username
Login to AoPS