Difference between revisions of "Ptolemy's Theorem"
m (Ptolemy's heorem moved to Ptolemy's Theorem: making a proper noun) |
|||
| Line 6: | Line 6: | ||
<math>ac+bd=ef</math>. | <math>ac+bd=ef</math>. | ||
| + | |||
| + | === Proof: Method I=== | ||
| + | |||
| + | 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.</math> Hence, <math>\displaystyle \angle ADP=\angle ABC</math> and <math>\displaystyle \triangle ABC \sim\triangle ADP</math> by <math>\displaystyle AA </math> similarity. | ||
| + | |||
| + | |||
| + | |||
| + | ===Proof: Method II=== | ||
| + | |||
=== Example === | === Example === | ||
Revision as of 12:49, 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: Method I
Given cyclic quadrilateral 
extend 
to 
such that 
Since quadrilateral is cyclic, 
However, 
is also supplementary to 
Hence, 
and 
by 
similarity.
Proof: Method II
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.
