Mathwords logoMathwords

Ptolemy's Theorem — Definition, Formula & Examples

Ptolemy's Theorem states that for any quadrilateral inscribed in a circle, the product of its two diagonals equals the sum of the products of its two pairs of opposite sides.

If ABCDABCD is a cyclic quadrilateral with consecutive vertices on a circle, then AC⋅BD=AB⋅CD+AD⋅BCAC \cdot BD = AB \cdot CD + AD \cdot BC, where ACAC and BDBD are the diagonals and ABAB, BCBC, CDCD, DADA are the sides.

Key Formula

AC⋅BD=AB⋅CD+AD⋅BCAC \cdot BD = AB \cdot CD + AD \cdot BC
Where:
  • ACAC = Length of diagonal from vertex A to vertex C
  • BDBD = Length of diagonal from vertex B to vertex D
  • AB,CDAB, CD = Lengths of one pair of opposite sides
  • AD,BCAD, BC = Lengths of the other pair of opposite sides

How It Works

Label the four vertices of the cyclic quadrilateral in order around the circle. Identify the two diagonals and the four sides. Multiply the lengths of opposite side pairs (AB⋅CDAB \cdot CD and AD⋅BCAD \cdot BC), then add those two products. The result equals the product of the two diagonal lengths. The theorem works only when the quadrilateral is inscribed in a circle; for non-cyclic quadrilaterals, the equality becomes a strict inequality known as Ptolemy's inequality.

Worked Example

Problem: A cyclic quadrilateral ABCD has sides AB = 3, BC = 4, CD = 5, and DA = 6. Diagonal BD = 7. Find the length of diagonal AC.
Write Ptolemy's equation: Apply the theorem to quadrilateral ABCD.
AC⋅BD=AB⋅CD+AD⋅BCAC \cdot BD = AB \cdot CD + AD \cdot BC
Substitute known values: Plug in the given side and diagonal lengths.
AC⋅7=3⋅5+6⋅4=15+24=39AC \cdot 7 = 3 \cdot 5 + 6 \cdot 4 = 15 + 24 = 39
Solve for AC: Divide both sides by 7.
AC=397≈5.57AC = \frac{39}{7} \approx 5.57
Answer: AC=397≈5.57AC = \dfrac{39}{7} \approx 5.57

Why It Matters

Ptolemy's Theorem appears frequently in math competitions and olympiad geometry as a powerful tool for relating lengths in cyclic figures. It also provides an elegant proof of the addition formulas for sine and cosine when applied to a cyclic quadrilateral inscribed in a unit circle.

Common Mistakes

Mistake: Applying the theorem to a quadrilateral that is not inscribed in a circle.
Correction: Ptolemy's equality holds only for cyclic quadrilaterals. For a general quadrilateral, you get Ptolemy's inequality: AC⋅BD≤AB⋅CD+AD⋅BCAC \cdot BD \leq AB \cdot CD + AD \cdot BC, with equality if and only if the quadrilateral is cyclic.