## Ptolemy's Theorem

### Stefan Gössner

Department of Mechanical Engineering, University of Applied Sciences, Dortmund, Germany.

Keywords: Ptolemy's theorem; vectorial proof; cyclic quadrilateral; geometry;

## Introduction

In Euclidean geometry, Ptolemy's theorem is a relation between the four sides and two diagonals of a cyclic quadrilateral (a quadrilateral whose vertices lie on a common circle) [1].

With given side and diagonal lengths, Ptolemy's theorem of a cyclic quadrilateral states:

$pq = ac+bd \\,.$(1)

For a cyclic quadrilateral the product of the diagonal lengths is equal to the sum of the product of the length of the pairs of opposite sides.

## Vectorial Proof

The quadrilateral can be described by a loop closure of side vectors $\\bold a$, $\\bold b$, $\\bold c$, $\\bold d$ running counter-clockwise (Fig.1) [3].

Expressing the diagonal vector $\\bold p$ by the sides of its two triangles $\\bold p = \\bold a + \\bold b = -(\\bold c + \\bold d)$ and squaring leads us to

$p^2 = a^2 + 2\\bold a\\bold b + b^2 = c^2 + 2\\bold c\\bold d + d^2 \\,.$(2)

Both angles in the triangle corners opposite to diagonal $\\bold p$ sum up to 180° and thus equals $\\cos A = -\\cos C$. So we can write in vectorial notation [4]

$\\frac{\\bold a \\bold b}{ab} = -\\frac{\\bold c \\bold d}{cd} \\quad or \\quad (\\bold a \\bold b)cd + (\\bold c \\bold d)ab = 0\\,.$(3)

Resolving for the dot products $\\bold a \\bold b$ and $\\bold c \\bold d$ in equation (2) and introducing them in (3) gives

$(p^2 - a^2 - b^2)\\,cd + (p^2 - c^2 - d^2)\\,ab = 0$

On our way for isolating $p^2$ we get

$p^2(ab + bc) = (a^2 + b^2)cd + (c^2 + d^2)ab\\,.$

Herein the right hand side can be simplified by resolving the brackets and reordering

$p^2 = \\frac{(ac+bd)(ad+bc)}{ab + cd}\\,.$(4)

Otherwise expressing the diagonal vector $\\bold q$ by its two triangle sides $\\bold q = \\bold b + \\bold c = -(\\bold d + \\bold a)$ and squaring again gives

$q^2 = b^2 + 2\\bold b\\bold c + c^2 = a^2 + 2\\bold a\\bold d + d^2 \\,.$

Both angles in the triangle corners opposite to diagonal $\\bold q$ also sum up to 180° and equal to $\\cos B = -\\cos D$. Then using this fact in an analog manner leads us to

$q^2 = \\frac{(ab+cd)(ac+bd)}{ad + bc}\\,.$(5)

Now multiplying equations (4) and (5), reducing the fractions and applying the square root finally results in (1)

$pq = ac+bd \\,.$