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 ${e},{f}$:


Proof 1

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

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 $(AC)(BD)=(AB)(CD)+(AD)(BC)$.

Proof 2 (inversion)

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

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

\[\frac{r^2}{AD\cdot BD}\cdot AB + \frac{r^2}{BD\cdot CD}\cdot BC \geq \frac{r^2}{AD\cdot CD}\cdot AC.\]

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


2023 AIME I Problem 5

Square $ABCD$ is inscribed in a circle. Point $P$ is on this circle such that $AP \cdot CP = 56$, and $BP \cdot DP = 90$. What is the area of the square?

Solution: We may assume that $P$ is between $B$ and $C$. Let $PA = a$, $PB = b$, $PC = C$, $PD = d$, and $AB = s$. We have $a^2 + c^2 = AC^2 = 2s^2$, because $AC$ is a diagonal. Similarly, $b^2 + d^2 = 2s^2$. Therefore, $(a+c)^2 = a^2 + c^2 + 2ac = 2s^2 + 2(56) = 2s^2 + 112$. Similarly, $(b+d)^2 = 2s^2 + 180$.

By Ptolemy's Theorem on $PCDA$, $as + cs = ds\sqrt{2}$, and therefore $a + c = d\sqrt{2}$. By Ptolemy's on $PBAD$, $bs + ds = as\sqrt{2}$, and therefore $b + d = a\sqrt{2}$. By squaring both equations, we obtain

\[2d^2 = (a+c)^2 = 2s^2 + 112\] \[2a^2 = (b+d)^2 = 2s^2 + 180.\]

Thus, $a^2 = s^2 + 90$, and $d^2 = s^2 + 56$. Plugging these values into $a^2 + c^2 = b^2 + d^2 = 2s^2$, we obtain $c^2 = s^2 - 90$, and $b^2 = s^2 - 56$. Now, we can solve using $a$ and $c$ (though using $b$ and $d$ yields the same solution for $s$).

\[(\sqrt{s^2 + 90})(\sqrt{s^2 - 90}) = ac = 56\] \[(s^2 + 90)(s^2 - 90) = 56^2\] \[s^4 = 90^2 + 56^2 = 106^2\] \[s^2 = 106.\]

The answer is $\boxed{106}$.

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: $\frac{1}{AB}=\frac{1}{AC}+\frac{1}{AE}$.

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