Difference between revisions of "Ptolemy's Theorem"

Revision as of 20:00, 14 April 2010

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

Definition

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 $\angle BAC=\angle DAP.$

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 $\frac{AD}{AP}=\frac{BD}{CP}\implies CP=\frac{(AP)(BD)}{(AD)}.$

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

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: 1/AB = 1/AC + 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 1/a=1/b+1/c.

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:

$(AC)(BD) = 7(AD) + 22$ $(AC)^2 = (AD)^2 - 121$ $(BD)^2 = (AD)^2 - 4$

Solving gives $AD = 14$

@@ Line 37: / Line 37: @@
 === 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>
+<math>(AC)^2 = (AD)^2 - 121</math>
+<math>(BD)^2 = (AD)^2 - 4</math>
+Solving gives <math>AD = 14</math>
 == See also ==

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