Math

Ptolemy’s inequality

A quadrilateral in a circle and its three products as rectangles at one scale: AB·CD and BC·DA laid end to end match AC·BD exactly while D stays on the circle. Unpin D and move it inside - the equality goes and an inequality stays.

By Mykhailo Makarov

Actual application image

What happens

For a quadrilateral ABCD inscribed in a circle, the product of the diagonals equals the sum of the products of opposite sides: AB·CD + BC·DA = AC·BD. Move any vertex along the circle and the equality holds; take D off the circle and it becomes an inequality, the left side always larger. The rectangles show the three products at one scale; they are an image of the numbers, not a cutting and rearranging of the rectangles.

Equations used in the module

Equations
A, B, C, D in this order
AB·CD + BC·DA ≥ AC·BD
equality ⇔ ABCD is a convex quadrilateral inscribed in a circle
gap = AB·CD + BC·DA − AC·BD ≥ 0
a self-intersecting ABCD is not the quadrilateral the equality is about

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