Stewart's theorem

The set of Stewart is a set of Euclidean geometry that will be used when describing the geometry of a triangle. It can be used, the length of a line through the corner of a triangle to the opposite side you calculate. It was erected in 1746 by the Scottish mathematician Matthew Stewart (although he was already known to Archimedes probably ).

Definition

Given a triangle ( see picture) with the defining vertices A, B and C and the side lengths

Next, let M be a point on the track with

The set of Stewart then states:

If the fraction designated, then applies ( with )

And the theorem can also be formulated as follows:

Applications

The important set of Heron for the calculation of the area of ​​a triangle from its side lengths follows directly from the set of Stewart. The set of generalized Stewart was also the Dutch mathematician Oene Bottema for use on simplices and tetrahedra.

The set of Stewart includes the Pythagorean theorem. In the special case and he states namely:

And thus:

This situation is at any given right triangle with right angle at always produce by allowing it reflects on the Kathetengerade, making and mirror points and the triangle is isosceles.

Proof of the theorem

We may assume without loss of generality without that the triangle (see picture) is a geometric figure of the complex plane and it is in its particularity which lies straight line coincides with the real axis, and it is also true, ie, the vertex in the upper half-plane. Otherwise, you can this situation by applying a suitably chosen plane Kongruenzabbildungen always create. Since congruent figures always have the same size relationships, it is sufficient to prove the theorem for this special case.

This then can be divided into three steps, the following calculations to the proof of the theorem do.

(I ) Basic equations

There are using the complex absolute value function the following basic equations (see picture):

(II ) Derived equations

From (I) results in first:

And further using the real part function and taking into account the fact that and:

You multiply in the penultimate equation with left and right, in the last equation with left and right, is the sum of the left and right terms and obtains because lifting off to the following sum representation:

(III ) Final equations

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

And finally because:

And thus, the above claimed identity (*).