Difference between revisions of "Ptolemy's Theorem"

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 $\displaystyle ABCD,$ extend $\displaystyle CD$ to $\displaystyle P$ such that $\angle BAC=\angle DAP.$

Since quadrilateral $\displaystyle ABCD,$ is cyclic, $\displaystyle m\angle ABC+m\angle ADC=180^\circ .$ However, $\displaystyle \angle ADP$ is also supplementary to $\angle ADC,$ so $\displaystyle \angle ADP=\angle ABC$ Hence, $\displaystyle \triangle ABC \sim \triangle ADP$ 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 $\angle BAC+\angle CAD=\angle DAP+\angle CAD \implies \angle BAC=\angle CAP,$ so $\triangle BAD\sim \triangle CAP.$ This yields $\frac{AD}{AP}=\frac{BD}{CP}\implies CP=\frac{(AP)(BD)}{(AD)}.$

However, $\displaystyle CP= CD+DP.$ Substituting in our expressions for $\displaystyle CP$ and $\displaystyle DP,$ $\frac{(AC)(BD)}{(AB)}=CD+\frac{(AD)(BC)}{(AB)}.$ Multiplying by $\displaystyle AB$ yields $\displaystyle (AC)(BD)=(AB)(CD)+(AD)(BC)$

--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.