# Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 36/refcontrol

    Triangles
In this and the next lecture we are dealing with triangles. A triangle is just a tuple $(A,B,C)$ of three points (called vertices) in an affine space $E$ (typically, an affine plane) over a Euclidean space $V$. We allow the situation that some or all vertices coincide (a degenerate triangle), and we identify triangles if they only differ by the ordering of the vertices. A triangle is by definition not degenerate if and only if the three points are affinely independent.MDLD/affinely independent Quite often, we regard a triangle as its convex hull, that is, the set
    ${\left\{rA+sB+tC\mid r+s+t=1,\,0\leq r,s,t\leq 1\right\}}$
of all barycentric combinationsMDLD/barycentric combinations of the three points, where all coefficients are not negative. The connecting segment
    ${\overline {A,B}}={\left\{rA+sB\mid r+s=1,\,0\leq r,s\leq 1\right\}}\,$
is called the edge between the vertices $A$ and $B$ (or the edge opposite of $C$). It is often denoted by $c$, its length is
    $d(A,B)=\Vert {\overrightarrow {AB}}\Vert \,.$
The corresponding denominations hold for the other edges. Sometimes we also denote the lengths of the edges by $a,b,c$. The angle $\angle (A,B,C)$ of the triangle at the vertex $B$ is defined as
    $\angle ({\overrightarrow {BA}},{\overrightarrow {BC}}),$
accordingly for the other vertices. The angles are usually denoted by $\alpha ,\beta ,\gamma$.

## Definition

Two trianglesMDLD/triangles in a Euclidean planeMDLD/Euclidean plane are called congruent if they can be transformed into each other by the compositionMDLD/composition of a translationMDLD/translation and an
We also say that congruent triangles can be transformed into each other by an affine-linear isometry.

## TheoremCreate referencenumber

Two triangles in a Euclidean planeMDLD/Euclidean plane are congruentMDLD/congruent (triangle)

### Proof

Translations and isometries preserve lengths; therefore, congruent triangles have the same lengths of edges. Let now two triangles $(A,B,C)$ and $(A',B',C')$ be given, having the same edge lengths. After renaming, we may assume that the edge lengths fulfill the relation
    $d(A,B)\geq d(A,C)\geq d(B,C)\,,$
and that the same holds for the second triangle. We can assume $E=\mathbb {R} ^{2}$, and we can also assume after a translation that $A=A'=0$ holds. After rotations of the triangles around the origin, we may further assume that $B$ as well as $B'$ lie on the positive $x$-axis. Because of the length equality, we have $B=B'$. The points $C$ and $C'$ have, on one hand, the same distance to $0$, and, on the other hand, the same distance to $B$. That is, they lie on the intersection of a circle about $0$ and a circle about $B$. Since there are only two intersection points, we have either $C=C'$, or $C$ and $C'$ can be transformed to each other by a reflection at the $x$-axis.
$\Box$

## Definition

Two trianglesMDLD/triangles in a Euclidean planeMDLD/Euclidean plane are called properly congruent if they can be transformed into each other by the compositionMDLD/composition of a translationMDLD/translation and a

## Definition

Two trianglesMDLD/triangles in a Euclidean planeMDLD/Euclidean plane are called similar if they can be transformed into each other by the compositionMDLD/composition of a translationMDLD/translation and an

## TheoremCreate referencenumber

Two trianglesMDLD/triangles in a Euclidean planeMDLD/Euclidean plane are similarMDLD/similar (triangle)

### Proof

$\Box$
    The Pythagorean theorem

## Definition

A triangleMDLD/triangle $(A,B,C)$ is called a right triangle if for one vertex the adjacent edges are perpendicularMDLD/perpendicular (R)

## Definition

In a right triangle,MDLD/right triangle the side opposite to the right angle is called the hypotenuse

## Definition

In a right triangle,MDLD/right triangle the sides adjacent to the right angle are called the catheti (or legs)
The Pythagorean theorem for a right triangle says the following.

## TheoremCreate referencenumber

In a right triangle,MDLD/right triangle the area of the square of the hypotenuseMDLD/hypotenuse

### Proof

We denote the vertices of the triangle by $A,B,C$, with the right angle located at $C$. We set $v=C-A$, and $w=B-C$. The vector from $A$ to $B$ equals $v+w$, and $v$ and $w$ are perpendicular to each other. Therefore,
    $\Vert {v+w}\Vert ^{2}=\left\langle v+w,v+w\right\rangle =\left\langle v,v\right\rangle +2\left\langle v,w\right\rangle +\left\langle w,w\right\rangle =\Vert {v}\Vert ^{2}+\Vert {w}\Vert ^{2}\,.$
$\Box$
In this formulation, we use that the area of a square (that is, of the geometric object) equals the (arithmetic) square of the side lengths. The proof does not refer to areas.

## Definition

For a triangleMDLD/triangle $A,B,C$ in a euclidean plane,MDLD/euclidean plane the line running through $A$ and being perpendicularMDLD/perpendicular to the line given by $B$ and $C$, is called the altitude line through $A$. The line segment from $A$ to the line given by $B$ and $C$ is called the altitude
$A$
The length of the altitude is also called the altitude.

## Definition

In a triangleMDLD/triangle $A,B,C$ in a euclidean plane,MDLD/euclidean plane the intersecting point of the altitude lineMDLD/altitude line through $A$ with the line given by $B$ and $C$ is called the foot
Der following theorem is called Cathetus theorem.

## TheoremCreate referencenumber

Let $A,B,C$ be a right triangle,MDLD/right triangle with the right angle located at $C$. Let $h$ denote the altitudeMDLD/altitude (triangle) through $C$, and let $D$ denote the footMDLD/foot (altitude) of this altitude on the line given by $A$ and $B$. Then
    $d(A,C)^{2}=d(A,D)\cdot d(A,B)\,.$

### Proof

We set $v=C-A$, and $w=B-C$. The vector from $A$ to $B$ equals $v+w$. We write the foot of the altitudeMDLD/foot of the altitude as
    $D=A+\lambda (v+w)\,,$
with $\lambda \in \mathbb {R}$; the vector of the altitude has the form
    $h=C-D=A+v-A-\lambda (v+w)=v-\lambda (v+w)\,.$
The orthogonality condition for the altitude yields
    ${\begin{aligned}0&=\left\langle v-\lambda (v+w),v+w\right\rangle \\&={\left(1-\lambda \right)}\left\langle v,v\right\rangle -\lambda \left\langle w,w\right\rangle \\&=\left\langle v,v\right\rangle -\lambda {\left(\left\langle v,v\right\rangle +\left\langle w,w\right\rangle \right)};\end{aligned}}$
therefore,
    $\lambda ={\frac {\left\langle v,v\right\rangle }{\left\langle v,v\right\rangle +\left\langle w,w\right\rangle }}\,.$
Hence, we obtain
    ${\begin{aligned}d(A,D)\cdot d(A,B)&=\lambda \Vert {v+w}\Vert \cdot \Vert {v+w}\Vert \\&=\lambda \left\langle v+w,v+w\right\rangle \\&=\lambda {\left(\left\langle v,v\right\rangle +\left\langle w,w\right\rangle \right)}\\&=\left\langle v,v\right\rangle \\&=\Vert {v}\Vert ^{2}\\&=d(A,C)^{2}.\end{aligned}}$
$\Box$
The following theorem is called Right triangle altitude theorem.

## TheoremCreate referencenumber

Let $A,B,C$ be a right triangle,MDLD/right triangle with the right angle located at $C$. Let $h$ be the altitudeMDLD/altitude (triangle) through $C$, and let $D$ denote the footMDLD/foot of this altitude through the linen given by $A$ and $B$. Then
    $d(C,D)^{2}=d(A,D)\cdot d(B,D)\,.$

### Proof

$\Box$
The following theorem is called Law of cosines.

## TheoremCreate referencenumber

Let a triangleMDLD/triangle $(A,B,C)$ be given, with side lengths $a,b,c$. Let $\gamma$ denote the angleMDLD/angle at $C$. Then
    $c^{2}=a^{2}+b^{2}-2ab\cos \gamma \,$

### Proof

$\Box$
    Thales's theorem

## TheoremCreate referencenumber

Let $P$ be a point in the Euclidean planeMDLD/Euclidean plane $E$, let $K$ be a circle with center $P$, and let $G$ be a line through $P$, intersecting the circle in the points $A$ and $B$. Then for every point $C\in K$,
$A,B,C$
$C$

### Proof

Without loss of generality, we may assume that $E=\mathbb {R} ^{2}$ and $P=0$, we denote the radius by $r$. We write $w=C$, $v=A$, $B=-v$. The vector from $A$ to $C$ is $-v+w$, and the vector from $B$ to $C$ is $v+w$. Therefore,
    ${\begin{aligned}\left\langle -v+w,v+w\right\rangle &=\left\langle -v,v\right\rangle +\left\langle -v,w\right\rangle +\left\langle w,v\right\rangle +\left\langle w,w\right\rangle \\&=-\left\langle v,v\right\rangle -\left\langle v,w\right\rangle +\left\langle v,w\right\rangle +\left\langle w,w\right\rangle \\&=-\left\langle v,v\right\rangle +\left\langle w,w\right\rangle \\&=-r^{2}+r^{2}\\&=0.\end{aligned}}$
This means that the two sides of the triangle meeting at $C$ are perpendicular to each other.
$\Box$
    The intercept theorems
We formulate in the language of linear algebra the intercept theorems. We work in the setting of a two-dimensional Euclidean vector space, which provides lengths of line segments. Two affine linen are called parallel if they are spanned by the same vector. We formulate the intercept theorems in such a way that the intersecting point of the rays lies in the origin. This can always be achieved by translating the intersecting point to the origin, as this does not change the lengths.

## TheoremCreate referencenumber

Let $V$ denote a two-dimensional Euclidean vector space,MDLD/Euclidean vector space and let $s_{1},s_{2},v\in V$ be vectors and suppose that $v$ and $s_{1}$ are linearly independent,MDLD/linearly independent and that $v$ and $s_{2}$ are linearly independent. Let $S_{1}=\mathbb {R} s_{1}$ and $S_{2}=\mathbb {R} s_{2}$ denote the lines (the rays) defined by $s_{1}$ and $s_{2}$. Let $P$ and $Q$ be points in $V$ with the corresponding parallel lines $G=P+\mathbb {R} v$ and $H=Q+\mathbb {R} v$. We denote the intersecting points (which exist uniquely due to the condition) by
    $A_{1}=S_{1}\cap G,\,A_{2}=S_{2}\cap G,\,B_{1}=S_{1}\cap H,\,B_{2}=S_{2}\cap H,$
and suppose that $A_{1},B_{1}\neq 0$. Then
    ${\frac {d(B_{1},B_{2})}{\Vert {B_{1}}\Vert }}={\frac {d(A_{1},A_{2})}{\Vert {A_{1}}\Vert }}\,.$

### Proof

Without loss of generality, we may assume that $s_{1}=A_{1}$, $s_{2}=A_{2}$, and $v=A_{2}-A_{1}$, as this does not change the lines involved. We write $B_{1}=tA_{1}$. We have $A_{2}=A_{1}+v$; therefore, we obtain
    $tA_{2}=tA_{1}+tv=B_{1}+tv\,.$
This point belongs to $S_{2}$ and also to $H$. This means that this point is just $B_{2}$. Hence, $B_{2}=tA_{2}$, and
    ${\frac {\Vert {B_{1}-B_{2}}\Vert }{\Vert {B_{1}}\Vert }}={\frac {\Vert {tA_{1}-tA_{2}}\Vert }{\Vert {tA_{1}}\Vert }}={\frac {\Vert {A_{1}-A_{2}}\Vert }{\Vert {A_{1}}\Vert }}\,.$
$\Box$
In particular, the preceding theorem says that in the given situation, the ratios of corresponding side lengths of the triangles $0,A_{1},A_{2}$ and $0,B_{1},B_{2}$ are the same. These triangles are similar; the triangle $0,B_{1},B_{2}$ arises from the triangle $0,A_{1},A_{2}$ by a homothety with factor $t$. This factor describes all length relations.

## CorollaryCreate referencenumber

Let $V$ denote a two-dimensional Euclidean vector space,MDLD/Euclidean vector space and let $s_{1},s_{2},v\in V$ be vectors and suppose that $v$ and $s_{1}$ are linearly independent,MDLD/linearly independent and that $v$ and $s_{2}$ are linearly independent. Let $S_{1}=\mathbb {R} s_{1}$ and $S_{2}=\mathbb {R} s_{2}$ denote the lines defined by $s_{1}$ and $s_{2}$. Let $P$ and $Q$ be points in $V$ with the corresponding parallel lines $G=P+\mathbb {R} v$ and $H=Q+\mathbb {R} v$. We denote the intersecting points (which exist uniquely due to the condition) by
    $A_{1}=S_{1}\cap G,\,A_{2}=S_{2}\cap G,\,B_{1}=S_{1}\cap H,\,B_{2}=S_{2}\cap H,$
and suppose that $A_{1},A_{2},A_{1}-A_{2}\neq 0$. Then
    ${\frac {d(B_{1},B_{2})}{d(A_{1},A_{2})}}={\frac {\Vert {B_{1}}\Vert }{\Vert {A_{1}}\Vert }}={\frac {\Vert {B_{2}}\Vert }{\Vert {A_{2}}\Vert }}\,.$

### Proof

This follows directly from Fact *****.
$\Box$
In particular, we have the equation
    ${\frac {\Vert {B_{1}}\Vert }{\Vert {A_{1}}\Vert }}={\frac {\Vert {B_{2}}\Vert }{\Vert {A_{2}}\Vert }}\,,$
which only takes the lengths on the rays into account.
In the last variant of the intercept theorem, there are three rays.

## TheoremCreate referencenumber

Let $V$ be a two-dimensional Euclidean vector space,MDLD/Euclidean vector space let $s_{1},s_{2},s_{3}\in V$ be vectors, and suppose that $v\in V$ is linearly independentMDLD/linearly independent to each of these vectors. Let $S_{i}=\mathbb {R} s_{i}$, $i=1,2,3$, be the lines defined by the $s_{i}$. Let $P$ and $Q$ be points in $V$ with den corresponding parallel lines $G=P+\mathbb {R} v$ and $H=Q+\mathbb {R} v$. We denote the intersecting points of the lines (which exist uniquely due to the conditions) by
    $A_{1}=S_{1}\cap G,\,A_{2}=S_{2}\cap G,\,A_{3}=S_{3}\cap G,\,B_{1}=S_{1}\cap H,\,B_{2}=S_{2}\cap H,\,B_{3}=S_{3}\cap H,$
and suppose that $B_{1}\neq B_{2}$. Then
    ${\frac {d(B_{2},B_{3})}{d(A_{2},A_{3})}}={\frac {d(B_{1},B_{2})}{d(A_{1},A_{2})}}\,.$

### Proof

We apply Fact ***** twice, to the situation given by $s_{1},s_{2}$ and to the situation given by $s_{2},s_{3}$, and we obtain
    ${\frac {d(B_{1},B_{2})}{d(A_{1},A_{2})}}={\frac {\Vert {B_{2}}\Vert }{\Vert {A_{2}}\Vert }}={\frac {d(B_{2},B_{3})}{d(A_{2},A_{3})}}\,.$
$\Box$