TheoremProved

Ptolemy's theorem

Statement

For any cyclic quadrilateral $ABCD$ (with vertices in order around a circle), the product of the lengths of the diagonals equals the sum of the products of the lengths of opposite sides: $AC \cdot BD = AB \cdot CD + BC \cdot AD$ .

Why is it true?

When a rectangle is inscribed in a circle, its diagonals are diameters of length $c$ and its opposite sides are pairs of equal legs $a$ and $b$ , so $AC \cdot BD = AB \cdot CD + BC \cdot AD$ reduces directly to the Pythagorean theorem $c^2 = a^2 + b^2$ . Deforming the rectangle by sliding its four vertices along the same circle preserves the inscribed angles subtending each arc, which locks the triangles formed by the sides and diagonals into similar pairs whose side ratios still add up to the diagonal product.

Proof sketch

Choose the point $M$ on the diagonal $BD$ such that $\angle BAM = \angle CAD$ . Since the inscribed angles $\angle ABM$ and $\angle ACD$ subtend the same arc $AD$ , they are equal, making $\triangle ABM$ similar to $\triangle ACD$ ; hence $\dfrac{AB}{AC} = \dfrac{BM}{CD}$ , or $AB \cdot CD = AC \cdot BM$ . Adding $\angle MAC$ to $\angle BAM = \angle CAD$ gives $\angle BAC = \angle MAD$ , and since $\angle BCA = \angle BDA$ , $\triangle ABC$ is similar to $\triangle AMD$ , giving $\dfrac{BC}{MD} = \dfrac{AC}{AD}$ , or $BC \cdot AD = AC \cdot MD$ . Summing the two equations yields $AB \cdot CD + BC \cdot AD = AC(BM + MD) = AC \cdot BD$ .

Topics that use this theorem

Quadrilaterals and polygons

Related theorems

Inscribed angle theorem Pythagorean theorem Law of sines

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

Claudius Ptolemy (translated by G. J. Toomer) (1998). Ptolemy's Almagest (Book I, Chapter 10)
H. S. M. Coxeter, S. L. Greitzer (1967). Geometry Revisited · DOI:10.5948/UPO9780883859346