Ptolemy's theorem: Difference between revisions

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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.

Statement

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

$ac+bd=ef.$

Proof

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

Problems

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 $A$ .

Solution

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: