Difference between revisions of "Ptolemy's Theorem"
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<math>ac+bd=ef</math>. | <math>ac+bd=ef</math>. | ||
| − | === Proof | + | === Proof === |
Given cyclic quadrilateral <math>\displaystyle ABCD,</math> extend <math>\displaystyle CD</math> to <math>\displaystyle P</math> such that <math>\angle BAC=\angle DAP.</math> | Given cyclic quadrilateral <math>\displaystyle ABCD,</math> extend <math>\displaystyle CD</math> to <math>\displaystyle P</math> such that <math>\angle BAC=\angle DAP.</math> | ||
| − | Since quadrilateral <math>\displaystyle ABCD,</math> is cyclic, <math>\displaystyle m\angle ABC+m\angle ADC=180^\circ .</math> However, <math>\displaystyle \angle ADP</math> is also supplementary to <math>\displaystyle \angle ADC | + | Since quadrilateral <math>\displaystyle ABCD,</math> is cyclic, <math>\displaystyle m\angle ABC+m\angle ADC=180^\circ .</math> However, <math>\displaystyle \angle ADP</math> is also supplementary to <math>\displaystyle \angle ADC,</math> so <math>\displaystyle \angle ADP=\angle ABC</math> Hence, <math>\displaystyle \triangle ABC \sim \triangle ADP</math> by <math>\displaystyle AA </math> similarity and <math>\displaystyle \frac{AB}{AD}=\frac{BC}{DP}\implies DP=\frac{(AD)(BC)}{(AB)}.</math> |
| + | Now, note that <math>\displaystyle \angle ABD=\angle ACD </math> (subtend the same arc) and <math>\displaystyle \angle BAC+\angle CAD=\angle DAP+\angle CAD \implies \angle BAC=\andle CAP,</math> so <math>\displaystyle \triangle BAD\sim \triangle CAP.</math> This yields <math>\displaystyle \frac{AD}{AP}=\frac{BD}{CP}\implies CP=\frac{(AP)(BD)}{(AD)}.</math> | ||
| + | However, <math>\displaystyle CP= CD+DP.</math> Substituting in our expressions for <math>\displaystyle CP</math> and <math>\displaystyle DP,</math> <math>\displaystyle \frac{(AC)(BD)}{(AB)}=CD+\frac{(AD)(BC)}{(AB)}.</math> Multiplying by <math>\displaystyle AB</math> yields <math>\displaystyle (AC)(BD)=(AB)(CD)+(AD)(BC)</math> | ||
| − | + | --[[User:4everwise|4everwise]] 14:09, 22 June 2006 (EDT) | |
Revision as of 13:09, 22 June 2006
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 
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 
similarity and 
Now, note that 
(subtend the same arc) and $\displaystyle \angle BAC+\angle CAD=\angle DAP+\angle CAD \implies \angle BAC=\andle CAP,$ (Error compiling LaTeX. Unknown error_msg) so 
This yields 
However, 
Substituting in our expressions for 
and 
Multiplying by 
yields 
--4everwise 14:09, 22 June 2006 (EDT)
Example
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; and 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.
