Difference between revisions of "Ptolemy's Theorem"
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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> | 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{ | + | 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>. | 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>. | ||
Revision as of 13:51, 6 June 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.
Contents
Definition
Given a cyclic quadrilateral 
with side lengths 
and diagonals 
:
.
Proof
Given cyclic quadrilateral 
extend 
to 
such that 
Since quadrilateral is cyclic, 
However, 
is also supplementary to 
so 
. Hence, 
by AA similarity and 
Now, note that 
(subtend the same arc) and 
so 
This yields 
However, 
Substituting in our expressions for 
and 
Multiplying by 
yields 
.
Problems
Equilateral Triangle Identity
Let 
be an equilateral triangle. Let be a point on minor arc of its circumcircle. Prove that 
.
Solution: Draw 
, 
, 
. By Ptolemy's Theorem applied to quadrilateral 
, we know that 
. Since 
, we divide both sides of the last equation by 
to get the result: .
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 
and the sixth, denoted by 
, has length 
. Find the sum of the lengths of the three diagonals that can be drawn from 
.
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 , 
being the diameter. 
, 
, and 
. Construct diagonals 
and 
. Notice that these diagonals form right triangles. You get the following system of equations:
(Ptolemy's Theorem)
$\n(AC)^2 = (AD)^2 - 121$ (Error compiling LaTeX. Unknown error_msg)
Solving gives 
