Ptolemy's Theorem

by

Jisook Oh

Let a quadrilateral ABCD be inscribed in a circle. Then the sum of the products of the two pairs of opposite sides equal the product of its two diagonals. In other words,

AB*DC+AD*BC=AC*BD

1. Proof I

Refer to the diagram at the below. The sides of the quadrilateral are chords of the circle, so the angles that each subtends at points on the circumference (such as A,B,C, and D) are equal. The angles marked with green dot and yellow dot are equal for this reason. Ptolemy's proof uses the line BE drawn so that the angles marked with pinks dot are equal.

Figure 1 Figure 2 Figure 3

First, consider triangle ABE and triangle DBC.The point E divides the diagonal AC so that AE+EC=AC. The triangles ABE and DBC are similar by angle-angle similarity. since sides in similar triangles are proportional,

AE/DC=AB/BD.

Next, consider triangle CBE and DBA.Since angles inscribing a common arc AB of circle are congruent, angle ADB and angle ACB with yellow dot are same. By angle-angle similarity, the triangle CBE and DBA are also similar, so that

EC/AD=BC/BD.

From these two equalities, we have

AE*BD=AB*DC

EC*BD=BC*AD.

By adding them, we get

(AE+EC)*BD=AB*DC+BC*AD.

Since AE+EC=AC,

AB*DC+AD*BC=AC*BD.

2. Proof II

This is the proof using the trigonometry. Refer the diagram of the below.What we need to prove is ac+bd=mn.

 

Let x=A/2, y=C/2, w=D/2. By the lemma , a=2r sinx, b=2r siny, c=2r sinz, d=2r sinw.

With this notation, the equation we want to prove becomes

(2r sinx)(2r sinz) + (2r siny)(2r sinw) = 2r sin(y+z) 2r sin (x+y)

2(sinx)(sinz) + 2(siny)(sinw) = 2 sin(y+z) sin(x+y)

 

Since A+B+C+D=360'

A/2+B/2+C/2+D/2=180'

(x+y+z)+w=180,

then sinw= sin(x+y+z).

2(sinx)(sinz) + 2(siny)(sin(x+y+z)) = 2 sin(y+z) sin(x+y).

We use the trig identity

2 (sinu)(sinv) = cos(u-v) - cos (u+v)

Using this identity, the equation becomes

cos(x-z) - cos(x+z) + cos(y-(x+y+z)) - cos(y+(x+y+z))= cos((y+z) - (x+y)) - cos((y+z) + (x+y))

cos(x-z) - cos(x+z) + cos(-x-z) - cos(x+2y+z) = cos(z-x) - cos(x+2y+z)

cos(x-z) - cos(x+z) + cos(x+z) - cos(x+2y+z) = cos(x-z) - cos(x+2y+z)

which is true for all x,y, and z.

3. Somethig from Ptolemy's theorem.

If the quadrilateral is a rectangle, the Pythagorean theorem follows at once, because the opposite sides are the sides of right triangles, and the diagonals, which are diameters of the circle, are the hypotenuses.Here is the GSP file. Verify this for yourself!

 

If the circle is assumed to have unit diameter, then the chords are equal to the sines of the angles they subtend at the circumference.By taking special quadrilaterals important trigonometric identities are obtained. For example, if one diagonal is a diameter, then it subtends right angles at the circumference, so that the sides of the quadrilateral are the snes and cosines of the two acute angles of the right triangles, and the other diagonal is the sine of the sum of the acute angles.Hence, we get the addition formula sine.

4. One Problem using Ptolemy's theorem.

Let ABC denote an equilateral triangle inscribed in a circle.

 

For any point P on the circle, show that the two shorter segments among PA, PB, PC add up to the third one.

Solution

Let s denote the length of the side of the given triangle. By Ptolemey's Theorem we have

s*PA = s*PB + s*PC

Therefore

PA = PB + PC.

Click here to explore this problem on the GSP file

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