Ptolemy's Theorem

Diagonals and sides of a cyclic quadrilateral: pq = ac + bd.

The idea

Theorem (Ptolemy's theorem).

Let $ABCD$ be a cyclic quadrilateral — four points taken in order around a circle. Then the product of its two diagonals equals the sum of the products of its two pairs of opposite sides: $AC \cdot BD = AB \cdot CD + AD \cdot BC.$

Naming the sides in order $a = AB$, $b = BC$, $c = CD$, $d = DA$ and the diagonals $p = AC$, $q = BD$, the same statement reads $pq = ac + bd$.

The equation ties the six lengths of a cyclic quadrilateral together: fixing any five of them determines the sixth. It also holds only on a circle. For four points taken in order that do not all lie on one circle, the strict inequality $AC \cdot BD < AB \cdot CD + AD \cdot BC$ holds instead, so equality occurs exactly when the four points lie on one circle.

Ways to work on it

• Walkthrough. State Ptolemy's theorem and see the rectangle case recover Pythagoras.
• Proof. Prove Ptolemy's theorem by constructing a point on a diagonal and pairing similar triangles.
• Practice. Solve for an unknown diagonal of a cyclic quadrilateral.
• Hardest. Apply Ptolemy to a regular pentagon to derive the golden ratio.

Not sure where to start? Take the ten-question placement test.