# Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 37/refcontrol

Beside the three vertices of a triangle, there are further characteristic points of a triangle, as its centroid (center of gravity), the circumcenter, den Incenter and the orthocenter.
    Medians and centroid

## Definition

For a set of $n$ points $P_{1},\ldots ,P_{n}$ in an affine spaceMDLD/affine space $E$ over a real vector space $V$, the barycentric combinationMDLD/barycentric combination
    ${\frac {1}{n}}P_{1}+{\frac {1}{n}}P_{2}+\cdots +{\frac {1}{n}}P_{n}$
is called the center of gravity
This is also called the centroid of the points. It just the barycentric combination of the points where every point enters with the same weight. For two points $P,Q\in E$, the center of gravity ${\frac {1}{2}}P+{\frac {1}{2}}Q$ is also called the midpoint of the two points (or of the line segment ${\overline {P,Q}}$). In the case of two real numbers, we also talk about the arithmetic mean of the numbers. For $n=1,2,3$, the centroid of the points is also the centroid of its convex hull. The centroid of the three points is also the intersection point of the medians of the triangle.

## Definition

For a nondegenerate triangleMDLD/nondegenerate triangle $A,B,C$ in a Euclidean plane,MDLD/Euclidean plane the line
    ${\frac {B+C}{2}}+s{\left(A-{\frac {B+C}{2}}\right)},\,s\in \mathbb {R} ,$
is called the median
$A$
The median through $A$ runs through the point $A$ and the midpoint of the opposite side of the triangle.

## TheoremCreate referencenumber

In a nondegenerate triangleMDLD/nondegenerate triangle in the Euclidean plane,MDLD/Euclidean plane the three mediansMDLD/medians (triangle)
${\frac {A+B+C}{3}}$

### Proof

We consider the condition
    ${\frac {B+C}{2}}+s{\left(A-{\frac {B+C}{2}}\right)}={\frac {A+C}{2}}+t{\left(B-{\frac {A+C}{2}}\right)}\,,$
which yields
    ${\begin{aligned}{\frac {B-A}{2}}&=t{\left(B-{\frac {A+C}{2}}\right)}-s{\left(A-{\frac {B+C}{2}}\right)}\\&={\left(-s-{\frac {t}{2}}\right)}A+{\left(t+{\frac {s}{2}}\right)}B+{\frac {s-t}{2}}C.\end{aligned}}$
We may assume $E=\mathbb {R} ^{2}$ and $A=0$. This implies, because $B$ and $C$ are linearly independent,MDLD/linearly independent the identity
    $s=t\,.$
Therefore,
    ${\frac {B-A}{2}}={\left(-{\frac {3}{2}}s\right)}A+{\left({\frac {3}{2}}s\right)}B\,,$
and we obtain
    $s={\frac {1}{3}}\,.$
Hence, the intersecting point is
    ${\frac {B+C}{2}}+{\frac {1}{3}}{\left(A-{\frac {B+C}{2}}\right)}={\frac {1}{3}}A+{\frac {1}{3}}B+{\frac {1}{3}}C\,.$
Due to the symmetry of the situation, this is also the intersecting point with the third median.
$\Box$
In particular, the centroid divides every median with the ratio $2:1$, the longer part being closer to the vertex.
    Perpendicular bisectors and circumcenter

## Definition

For two points $A\neq B$ in the Euclidean plane,MDLD/Euclidean plane the lineMDLD/line (affine) that is perpendicularMDLD/perpendicular to the line given by $A$ and $B$, and that runs through the midpointMDLD/midpoint (line segment) of the line segment between $A$ and $B$, is called the perpendicular bisector
The perpendicular bisector can be described as
    ${\frac {A+B}{2}}+s(B-A)^{\perp },\,s\in \mathbb {R} ,$
where $(B-A)^{\perp }$ is an arbitrary vector $\neq 0$ perpendicular to $B-A$. If $A={\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}}$ and $B={\begin{pmatrix}b_{1}\\b_{2}\end{pmatrix}}$ are given in Cartesian coordinates, then the perpendicular bisector is
    ${\frac {1}{2}}{\begin{pmatrix}a_{1}+b_{1}\\a_{2}+b_{2}\end{pmatrix}}+s{\begin{pmatrix}b_{2}-a_{2}\\-b_{1}+a_{1}\end{pmatrix}},\,s\in \mathbb {R} .$

## LemmaCreate referencenumber

Let $A,B$ be different points in a Euclidean plane.MDLD/Euclidean plane Then the perpendicular bisectorMDLD/perpendicular bisector of $A$ and $B$ consists of all those points that have the same distanceMDLD/distance (Euclidean) to
$A$
$B$

### Proof

$\Box$

## TheoremCreate referencenumber

The perpendicular bisectorsMDLD/perpendicular bisectors of the three sides in a nondegenerate triangleMDLD/nondegenerate triangle in the

### Proof

The perpendicular bisector for the line segment between $A$ and $B$ consists, according to Fact *****, exactly of those points of the plane that have to these points the same distance. Therefore, the intersecting point $P$ of the perpendicular bisector of $A$ and $B$ and the perpendicular bisector of $A$ and $C$ has the same distance to all three vertices. This gives the supplement, and also, that all three perpendicular bisectors meet in this point.
$\Box$

## Definition

The intersecting point of the three perpendicular bisectorsMDLD/perpendicular bisectors in a nondegnerate triangleMDLD/nondegnerate triangle in the Euclidean planeMDLD/Euclidean plane is called the
The circumcenter is the center of the circumcircle; this is the circle that contains all three vertices of the triangle.
    Angle bisectors and incenter

## Definition

For two linearly independentMDLD/linearly independent vectors $v$ and $w$ in a normedMDLD/normed (vs) real vector spaceMDLD/real vector space $V$, the line generated by
    ${\frac {v}{\Vert {v}\Vert }}+{\frac {w}{\Vert {w}\Vert }}$
is called the angle bisector
We define here the angle bisectors without referring to an angle, we do not even suppose that an inner product is given. However, if both vectors are in a Euclidean space, then an easy consideration (see exercise *****) show that the angle bisector cut indeed the angle into halves. The definition works also for an affine space over a normed vector space; three non-collinear points define an angle bisector for every point involved.

## LemmaLemma 37.9 change

Let $v,w$ be linearly independentMDLD/linearly independent vectors in $\mathbb {R} ^{2}$. Then every point of the angle bisectorMDLD/angle bisector of $v$ and $w$ has the same distanceMDLD/distance (Euclidean) to the lines $\mathbb {R} v$ and to $\mathbb {R} w$. If a point has the same distance to $\mathbb {R} v$ and to $\mathbb {R} w$, then it belongs to the angle bisector of $v$ and $w$ or to the angle bisector of
$v$
$-w$

### Proof

We may assume that $v$ and $w$ are normed.MDLD/normed (vector) Let $P\in \mathbb {R} ^{2}$. Due to Fact *****, we have
    $d(P,\mathbb {R} v)^{2}=\Vert {P}\Vert ^{2}-\left\langle P,v\right\rangle ^{2}\,,$
and, accordingly,
    $d(P,\mathbb {R} w)^{2}=\Vert {P}\Vert ^{2}-\left\langle P,w\right\rangle ^{2}\,.$
Therefore, the distances coincide if and only if
    $\left\langle P,v\right\rangle =\pm \left\langle P,w\right\rangle \,$
holds. If
    $P=s(v+w)\,,$
then
    $\left\langle P,v\right\rangle =\left\langle s(v+w),v\right\rangle =s+s\left\langle w,v\right\rangle =\left\langle s(v+w),w\right\rangle =\left\langle P,w\right\rangle \,,$
and the equation holds. For the reverse statement, we may write
    $P=sv+tw\,.$
From
    $\left\langle P,v\right\rangle =\left\langle P,w\right\rangle \,$
we can deduce
    $s+t\left\langle v,w\right\rangle =\left\langle sv+tw,v\right\rangle =\left\langle sv+tw,w\right\rangle =t+s\left\langle v,w\right\rangle \,;$
therefore,
    $s{\left(1-\left\langle v,w\right\rangle \right)}=t{\left(1-\left\langle v,w\right\rangle \right)}\,.$
Because $v$ and $w$ are normed and linearly independent, exercise ***** implies that
    $\vert {\left\langle v,w\right\rangle }\vert <1\,,$
hence, the right-hand factor is not $0$, and, therefore, $s=t$. In case
    $\left\langle P,v\right\rangle =-\left\langle P,w\right\rangle \,,$
a similar consideration gives $s=-t$.
$\Box$

## TheoremCreate referencenumber

The three angle bisectorsMDLD/angle bisectors in a nondegenerate triangleMDLD/nondegenerate triangle meet in one intersecting point, which has the same distance to every side of the triangle. If the vertices are given by $A={\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}},B={\begin{pmatrix}b_{1}\\b_{2}\end{pmatrix}},C={\begin{pmatrix}c_{1}\\c_{2}\end{pmatrix}}$, and if the side lengths are denoted by $a=d(B,C),b=d(A,C),c=d(A,B)$, then this intersecting point has the coordinates
    ${\frac {1}{a+b+c}}{\begin{pmatrix}aa_{1}+bb_{1}+cc_{1}\\aa_{2}+bb_{2}+cc_{2}\end{pmatrix}}.$

### Proof

Due to Lemma 37.9 , the angle bisector for $A$ consists of points that have the same distance to the line of the adjacent sides $\mathbb {R} {\overrightarrow {AB}}$ and $\mathbb {R} {\overrightarrow {AC}}$. Accordingly, the angle bisector for $B$ consists of points that have the same distance to the lines $\mathbb {R} {\overrightarrow {BA}}$ and $\mathbb {R} {\overrightarrow {BC}}$. Therefore, the intersecting point of these two angle bisectors has the same distance to all three sides. Moreover, the inner product of this intersecting point with the three normed vectors of the sides is always the same, as the proof of Lemma 37.9
shows. Therefore, again by Lemma 37.9 , this intersecting point lies also on the third angle bisector.
In order to determine the coordinates, we write the angle bisector of $A$ as
    $A+s{\left({\frac {\overrightarrow {AB}}{c}}+{\frac {\overrightarrow {AC}}{b}}\right)},$
that is,
    ${\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}}+s{\left({\frac {1}{c}}{\begin{pmatrix}b_{1}-a_{1}\\b_{2}-a_{2}\end{pmatrix}}+{\frac {1}{b}}{\begin{pmatrix}c_{1}-a_{1}\\c_{2}-a_{2}\end{pmatrix}}\right)}.$
Equating with the angle bisector of $B$, we obtain
    ${\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}}+s{\left({\frac {1}{c}}{\begin{pmatrix}b_{1}-a_{1}\\b_{2}-a_{2}\end{pmatrix}}+{\frac {1}{b}}{\begin{pmatrix}c_{1}-a_{1}\\c_{2}-a_{2}\end{pmatrix}}\right)}={\begin{pmatrix}b_{1}\\b_{2}\end{pmatrix}}+t{\left({\frac {1}{c}}{\begin{pmatrix}a_{1}-b_{1}\\a_{2}-b_{2}\end{pmatrix}}+{\frac {1}{a}}{\begin{pmatrix}c_{1}-b_{1}\\c_{2}-b_{2}\end{pmatrix}}\right)}\,.$
The solution is given by
    $s={\frac {bc}{a+b+c}}\,,$
and
    $t={\frac {ac}{a+b+c}}\,,$
because this gives, after inserting, the symmetric expression
    ${\begin{aligned}{\begin{pmatrix}a_{1}\\a_{2}\end{pmatrix}}+{\frac {bc}{a+b+c}}{\left({\frac {1}{c}}{\begin{pmatrix}b_{1}-a_{1}\\b_{2}-a_{2}\end{pmatrix}}+{\frac {1}{b}}{\begin{pmatrix}c_{1}-a_{1}\\c_{2}-a_{2}\end{pmatrix}}\right)}&={\frac {1}{a+b+c}}{\begin{pmatrix}a_{1}(a+b+c)+b(b_{1}-a_{1})+c(c_{1}-a_{1})\\a_{2}(a+b+c)+b(b_{2}-a_{2})+c(c_{2}-a_{2})\end{pmatrix}}\\&={\frac {1}{a+b+c}}{\begin{pmatrix}aa_{1}+bb_{1}+cc_{1}\\aa_{2}+bb_{2}+cc_{2}\end{pmatrix}}.\,\end{aligned}}$
This is also the intersecting point of all three angle bisectors.
$\Box$

## Definition

The intersecting point of the three angle bisectorsMDLD/angle bisectors in a nondegenerate triangleMDLD/nondegenerate triangle in the Euclidean planeMDLD/Euclidean plane is called
The circle around the incenter that meets all three sides of the triangles tangentially, is called incircle.
    The orthocenter

## LemmaLemma 37.12 change

Let $U$ be the circumcenter,MDLD/circumcenter and let $S$ be the centroidMDLD/centroid (triangle) of a nondegenerate triangleMDLD/nondegenerate triangle in the Euclidean plane.MDLD/Euclidean plane Then the point
    $U+3{\overrightarrow {US}}$
lies on every altitudeMDLD/altitude (triangle)

### Proof

We move $U$ to the origin, so that the points $A,B,C$ have the same norm. The point in question is just $A+B+C$. The vector of the line given by this point and $A$ is $B+C$. This line runs through $A$, and we have
    $\left\langle B+C,B-C\right\rangle =\left\langle B,B\right\rangle -\left\langle C,C\right\rangle \,.$
Due to the equality of the norms, this is $0$; therefore, this line is the altitude line through $A$.
$\Box$

## Definition

For a nondegenerate triangleMDLD/nondegenerate triangle $A,B,C$ in a Euclidean plane,MDLD/Euclidean plane the intersecting point of the three altitudesMDLD/altitudes (triangle) is called the
    The Euler line

## CorollaryCreate referencenumber

The centroid,MDLD/centroid (triangle) the circumcenter,MDLD/circumcenter and the orthocenterMDLD/orthocenter of a nondegenerate triangleMDLD/nondegenerate triangle in the

### Proof

This follows directly from Lemma 37.12 .
$\Box$
If the triangle is equilateral, then the three points coincide and there are many lines running through this point. If the triangle is not equilateral, then these points are not equal (see exercise *****) and there is exactly one line given by these three points. It is called the Euler line.
    The nine-point circle

## LemmaCreate referencenumber

Let a nondegenerate triangleMDLD/nondegenerate triangle $\triangle$ in the Euclidean planeMDLD/Euclidean plane with vertices $A,B,C$ be given. Let $F$ be the circumcircleMDLD/circumcircle of the midpointsMDLD/midpoints
* The radius of $F$ is half of the radius of the circumcircle of $\triangle$.
* The line segements between the orthocenterMDLD/orthocenter and the vertices are cut in halves by $F$.
* The feet of the altitudesMDLD/feet of the altitudes of $\triangle$ lie on $F$.

### Proof

* Let $U$ be the circumcenterMDLD/circumcenter of the triangle; we may assume that this point is the origin of a Cartesian coordinate system. We consider the point The distance between the midpoint $A'={\frac {1}{2}}{\left(B+C\right)}$ of the side connecting $B$ and $C$ and $U'$ is Since the norms of all vertices $A,B,C$ are equal due to the choice of $U$, it follows that $U'$ is the circumcenter of the triangle given by the midpoints of the original triangle, and that its radius is the half of the radius of the circumcircle.
    $U'={\frac {1}{2}}{\left(A+B+C\right)}\,.$
    $\Vert {{\frac {1}{2}}{\left(B+C\right)}-U'}\Vert =\Vert {{\frac {1}{2}}{\left(B+C\right)}-{\frac {1}{2}}{\left(A+B+C\right)}}\Vert ={\frac {1}{2}}\Vert {A}\Vert \,.$
* By Lemma 37.12 , $A+B+C$ is the orthocenter.MDLD/orthocenter (triangle) Therefore, the midpoint of the line segment between $A$ and the orthocenter equals The distance between this point and $U'$ is
    ${\frac {1}{2}}(A+B+C)+{\frac {1}{2}}A=A+{\frac {1}{2}}(B+C)\,.$
    $\Vert {{\frac {1}{2}}(A+B+C)-{\left(A+{\frac {1}{2}}(B+C)\right)}}\Vert =\Vert {-{\frac {1}{2}}A}\Vert ={\frac {1}{2}}\Vert {A}\Vert \,.$
* Note that the points constructed in (1) and (2) lie on the circle $F$ opposite to each other. Indeed, we have which is the center of $F$. Hence, for each side, its midpoint, the midpoint between the opposite vertex and the orthocenter, and the foot of the corresponding altitude form a right triangle. Its circle with the hypotenuse as diameter equals $F$.
    ${\frac {1}{2}}{\left({\frac {1}{2}}(B+C)\right)}+{\frac {1}{2}}{\left(A+{\frac {1}{2}}(B+C)\right)}={\frac {1}{2}}{\left(A+B+C\right)}\,,$
$\Box$
The circle in the preceding statement is called the Nine-point circle, or the Feuerbach circle.