Difference between revisions of "Ptolemy's Theorem"

m (proofreading)
m (proofreading)
Line 25: Line 25:
 
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.
 
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''.
+
Now, Ptolemy's Theorem states that ''ab + ac = bc'', which is equivalent to ''1/a=1/b+1/c''.
  
 
== See also ==
 
== See also ==
 
* [[Geometry]]
 
* [[Geometry]]

Revision as of 13:54, 26 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 $ABCD$ with side lengths $\displaystyle {a},{b},{c},{d}$ and diagonals $\displaystyle {e},{f}$:

$\displaystyle ac+bd=ef$.

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; 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.

See also

Invalid username
Login to AoPS