# Stewart's Theorem

The set of Stewart is a set of Euclidean geometry , the one in the description of the geometry triangle is used. It can be used to calculate the length of a line through the corner of a triangle to the opposite side. It was drawn up in 1746 by the Scottish mathematician Matthew Stewart (although it was probably already known to Archimedes ).

## definition

Construction for Stewart's theorem

A triangle is given (see picture) with the defining corner points A, B and C and the side lengths

${\ displaystyle a = {\ overline {CB}}}$; and .${\ displaystyle b = {\ overline {AC}}}$${\ displaystyle c = {\ overline {AB}}}$

Further, let M be a point on the line with ${\ displaystyle {\ overline {AB}}}$

${\ displaystyle x = {\ overline {AM}}}$; and .${\ displaystyle y = {\ overline {MB}}}$${\ displaystyle c = x + y \,}$

The Stewart's theorem then says:

(1) ${\ displaystyle xa ^ {2} + yb ^ {2} = \ left (x + y \ right) {\ overline {CM}} ^ {2} + xy ^ {2} + yx ^ {2} = c \ left ({\ overline {CM}} ^ {2} + xy \ right)}$

Is the fraction with referred to, then (with ) ${\ displaystyle {\ frac {\ overline {AM}} {\ overline {AB}}}}$${\ displaystyle \ phi \,}$${\ displaystyle \ phi ^ {\ prime} = 1- \ phi}$

${\ displaystyle x = \ phi c \,}$and ,${\ displaystyle y = \ phi ^ {\ prime} c}$

and the sentence can also be formulated as follows:

(2) ${\ displaystyle {\ overline {CM}} ^ {2} = a ^ {2} \ phi + b ^ {2} \ phi ^ {\ prime} -c ^ {2} \ phi \ phi ^ {\ prime} }$

## Applications

Heron's important theorem for calculating the area of a triangle from its side lengths follows directly from Stewart's theorem. Stewart's theorem was also generalized for application to simplexes and tetrahedra by the Dutch mathematician Oene Bottema .

The Stewart's theorem includes the Pythagorean theorem . In the special case and it says: ${\ displaystyle a = b}$${\ displaystyle x = y}$

${\ displaystyle a ^ {2} = {\ overline {CM}} ^ {2} + x ^ {2}}$

and thus:

${\ displaystyle {\ overline {CB}} ^ {2} = {\ overline {CM}} ^ {2} + {\ overline {MB}} ^ {2}}$

For a given right-angled triangle with a right angle at, this situation can always be generated by mirroring it straight at the cathetus , whereby and become mirror points and the triangle becomes an isosceles . ${\ displaystyle CMB}$${\ displaystyle M}$ ${\ displaystyle CM}$ ${\ displaystyle A}$${\ displaystyle B}$ ${\ displaystyle ABC}$

## Proof of the theorem

One may without loss of generality assume that the triangle (see Figure), a geometrical figure of the complex plane represents and, in particular , the straight line with the real axis coincides at the same time applies, that is, the corner point in the upper half-plane is located. Otherwise, this situation can always be created by using suitably chosen plane congruence maps. Since congruent figures always have the same size relationships, it is sufficient to prove the theorem for this special case. ${\ displaystyle ABC}$${\ displaystyle A = 0}$ ${\ displaystyle AB = AM = MB}$ ${\ displaystyle \ mathbb {R} \ subset \ mathbb {C}}$${\ displaystyle C \ in \ mathbb {H}}$ ${\ displaystyle C}$

The following calculations to prove the theorem can then be made in three steps.

### (I) basic equations

Using the complex absolute value function, the following basic equations exist (see figure): ${\ displaystyle z \ mapsto | z | = {\ sqrt {z \ cdot {\ overline {z}}}}}$

${\ displaystyle A = 0}$
${\ displaystyle {\ overline {AC}} = | C | = b}$
${\ displaystyle M = {\ overline {AM}} = x}$
${\ displaystyle {\ overline {MB}} = | MB | = y}$
${\ displaystyle B = {\ overline {AB}} = {\ overline {AM}} + {\ overline {MB}} = x + y = c}$
${\ displaystyle {\ overline {BC}} = | BC | = | C- (x + y) | = a}$
${\ displaystyle {\ overline {CM}} = | Cx |}$

### (II) Derived equations

From (I) we get:

${\ displaystyle {\ overline {CM}} ^ {2} = | Cx | ^ {2} = (Cx) \ cdot {\ overline {(Cx)}}}$

and further using the real part function and taking into account the fact that and : ${\ displaystyle z \ mapsto \ Re (z)}$${\ displaystyle x = {\ overline {x}} \ in \ mathbb {R}}$${\ displaystyle y = {\ overline {y}} \ in \ mathbb {R}}$

${\ displaystyle a ^ {2} = | (Cx) -y | ^ {2} = ((Cx) -y) \ cdot {\ overline {((Cx) -y)}} = | Cx | ^ {2 } + y ^ {2} -2 \ cdot y \ cdot \ Re (Cx) = {\ overline {CM}} ^ {2} + y ^ {2} -2 \ cdot y \ cdot \ Re (Cx)}$
${\ displaystyle b ^ {2} = | C | ^ {2} = C \ cdot {\ overline {C}} = (C-x + x) \ cdot {\ overline {((Cx) + x)}} = | Cx | ^ {2} + x ^ {2} +2 \ cdot x \ cdot \ Re (Cx) = {\ overline {CM}} ^ {2} + x ^ {2} +2 \ cdot x \ cdot \ Re (Cx)}$

In the penultimate equation, one multiplies left and right with , in the last equation left and right with , forms the sum of the respective left and right terms and, as it disappears, the following sum is obtained : ${\ displaystyle x}$${\ displaystyle y}$${\ displaystyle 2 \ cdot x \ cdot y \ cdot \ Re (Cx)}$

${\ displaystyle x \ cdot a ^ {2} + y \ cdot b ^ {2} = x \ cdot ({\ overline {CM}} ^ {2} + y ^ {2}) + y \ cdot ({\ overline {CM}} ^ {2} + x ^ {2})}$

### (III) Final equations

From (II) it follows by multiplying and swapping the terms and after factoring out :

${\ displaystyle x \ cdot a ^ {2} + y \ cdot b ^ {2} = {\ overline {CM}} ^ {2} \ cdot (x + y) + x \ cdot y \ cdot (x + y )}$

and finally because of : ${\ displaystyle c = x + y}$

${\ displaystyle x \ cdot a ^ {2} + y \ cdot b ^ {2} = c \ cdot ({\ overline {CM}} ^ {2} + x \ cdot y)}$

and thus the identity claimed above (1).

## literature

• N. Altshiller-Court: Stewart's Theorem . In: College Geometry: A Second Course in Plane Geometry for Colleges and Normal Schools . 2nd ed.Barnes and Noble, 1952
• O. Bottema: An extension of Stewart's formula . In: Elements of Mathematics , 34/1979, pp. 138–140, ( )
• O. Bottema: De formule van Stewart voor een viervlak . In: Nieuw Tijdschrift voor Wiskunde , 68 / 1980–81, pp. 79–83,
• Harold Scott MacDonald Coxeter , SL Greitzer: Timeless geometry . Klett, 1983, ISBN 3-12-983390-0
• György Hajós : Introduction to Geometry . BG Teubner Verlag, Leipzig (Hungarian: Bevezetés A Geometriába . Translated by G. Eisenreich [Leipzig, also editing]).
• Helmut Karzel , Hans-Joachim Kroll: History of Geometry since Hilbert . Scientific Book Society, Darmstadt 1988, ISBN 3-534-08524-8 .

## Individual evidence

1. Helmut Karzel , Hans-Joachim Kroll: History of geometry since Hilbert . Scientific Book Society, Darmstadt 1988, ISBN 3-534-08524-8 , pp. 96 .
2. ^ Based on Helmut Karzel , Hans-Joachim Kroll: History of Geometry since Hilbert . Scientific Book Society, Darmstadt 1988, ISBN 3-534-08524-8 , pp. 384 . where, however, the proof is purely vectorial without the use of complex numbers.