Ptolemy's Theorem

Stefan Gössner

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

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


Fig.1: Cyclic Quadrilateral

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.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 a\\bold a, b\\bold b, c\\bold c, d\\bold d running counter-clockwise (Fig.1) [3].

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

p2=a2+2ab+b2=c2+2cd+d2.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 p\\bold p sum up to 180° and thus equals cosA=cosC\\cos A = -\\cos C. So we can write in vectorial notation [4]

abab=cdcdor(ab)cd+(cd)ab=0.\\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 ab\\bold a \\bold b and cd\\bold c \\bold d in equation (2) and introducing them in (3) gives

(p2a2b2)cd+(p2c2d2)ab=0(p^2 - a^2 - b^2)\\,cd + (p^2 - c^2 - d^2)\\,ab = 0

On our way for isolating p2p^2 we get

p2(ab+bc)=(a2+b2)cd+(c2+d2)ab. 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

p2=(ac+bd)(ad+bc)ab+cd.p^2 = \\frac{(ac+bd)(ad+bc)}{ab + cd}\\,.(4)

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

q2=b2+2bc+c2=a2+2ad+d2.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 q\\bold q also sum up to 180° and equal to cosB=cosD\\cos B = -\\cos D. Then using this fact in an analog manner leads us to

q2=(ab+cd)(ac+bd)ad+bc.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.pq = ac+bd \\,.