OpenLLM-France/wikimedia huggingface dataset:
# Triangle geometry/Perpendicular bisector/Introduction/Section
## Definition

Line segment/Plane/Perpendicular bisector/Definition
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
    ${\frac {1}{2}}{\begin{pmatrix}a_{1}+b_{1}\\a_{2}+b_{2}\end{pmatrix}}+s{\begin{pmatrix}b_{2}-a_{2}\\-b_{1}+a_{1}\end{pmatrix}},\,s\in \mathbb {R} .$

## Lemma
### Proof

$\Box$

## Theorem
### Proof

$\Box$

## Definition

Triangle/Perpendicular bisector/Circumcenter/Definition
The circumcenter is the center of the circumcircle; this is the circle that contains all three vertices of the triangle.
# Line segment/Plane/Perpendicular bisector/Definition

For two points $A\neq B$ in the Euclidean plane, the line that is perpendicular to the line given by $A$ and $B$, and that runs through the midpoint of the line segment between $A$ and $B$, is called the perpendicular bisector of the line segment.
# Perpendicular bisector/Distance condition/Fact

Let $A,B$ be different points in a Euclidean plane.
perpendicular bisector of $A$ and $B$ consists of all those points that have the same distance to
$A$
$B$
# Triangle/Perpendicular bisector/Intersecting point/Fact

The perpendicular bisectors of the three sides in a nondegenerate triangle in the Euclidean plane
All vertices of the triangle have the same distance to this intersecting point.
# Triangle/Perpendicular bisector/Intersecting point/Fact/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.
# Triangle geometry/Euler line/Introduction/Section
## Corollary
### Proof

$\Box$
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.
# Triangle geometry/Nine-point circle/Introduction/Section
## Lemma
### Proof

$\Box$
The circle in the preceding statement is called the Nine-point circle, or the Feuerbach circle.
# Triangle geometry/Angle bisectors/Introduction/Section
## Definition

Rays/Angle bisector/Definition
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.

## Lemma
### Proof

$\Box$

## Theorem
### Proof

$\Box$

## Definition

Triangle/Angle bisectors/Incenter/Definition
The circle around the incenter that meets all three sides of the triangles tangentially, is called incircle.
# Rays/Angle bisector/Definition

For two linearly independent vectors $v$ and $w$ in a normed real vector space $V$, the line generated by
    ${\frac {v}{\Vert {v}\Vert }}+{\frac {w}{\Vert {w}\Vert }}$
is called the angle bisector of the two vectors.
# Angle bisector/Distance condition/Fact

Let $v,w$ be linearly independent vectors in $\mathbb {R} ^{2}$.
angle bisector of $v$ and $w$ has the same distance 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
$v$
$-w$
# Angle bisector/Distance condition/Fact/Proof

We may assume that $v$ and $w$ are normed. 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 $s=-t$.
# Triangle/Angle bisectors/Intersecting point/Fact

The three angle bisectors in a nondegenerate 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
    ${\frac {1}{a+b+c}}{\begin{pmatrix}aa_{1}+bb_{1}+cc_{1}\\aa_{2}+bb_{2}+cc_{2}\end{pmatrix}}.$
# Triangle/Angle bisectors/Intersecting point/Fact/Proof

Due to fact, 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 fact shows. Therefore, again by fact, 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.
# Triangle/Angle bisectors/Incenter/Definition

The intersecting point of the three angle bisectors in a nondegenerate triangle in the Euclidean plane is called incenter.
# Linear algebra (Osnabrück 2024-2025)/Angle bisector/Distance condition/Fact/Factreferencenumber

Lemma 37.9
# Circumcenter as origin/Sum of points on altitude/Fact

Let $U$ be the circumcenter, and let $S$ be the centroid of a nondegenerate triangle in the Euclidean plane.
    $U+3{\overrightarrow {US}}$
lies on every altitude
In particular, the three altitudes intersect in one point.
# Circumcenter as origin/Sum of points on altitude/Fact/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 $A$.
# Triangle/Orthocenter/Definition

For a nondegenerate triangle $A,B,C$ in a Euclidean plane, the intersecting point of the three altitudes is called the orthocenter.
# Triangle/Euler line/Existence/Fact

The centroid, the circumcenter, and the orthocenter of a nondegenerate triangle in the Euclidean plane
# Triangle/Euler line/Existence/Fact/Proof

This follows directly from fact.
# Linear algebra (Osnabrück 2024-2025)/Circumcenter as origin/Sum of points on altitude/Fact/Factreferencenumber

Lemma 37.12
# Triangle/Nine-point circle/Fact/Proof

* Let $U$ be the 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 fact, $A+B+C$ is the orthocenter. 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 $F$.
    ${\frac {1}{2}}{\left({\frac {1}{2}}(B+C)\right)}+{\frac {1}{2}}{\left(A+{\frac {1}{2}}(B+C)\right)}={\frac {1}{2}}{\left(A+B+C\right)}\,,$
# MDLD/medians (triangle)

Start=medians (triangle) See=median (triangle) MDLD/median (triangle) Target=[[MDLD/median (triangle)]]
# MDLD/median (triangle)

Start=median (triangle) Target=Triangle/Median/Definition
# MDLD/perpendicular bisector

Start=perpendicular bisector Target=Line segment/Plane/Perpendicular bisector/Definition
# MDLD/perpendicular bisectors

Start=perpendicular bisectors See=perpendicular bisector MDLD/perpendicular bisector Target=Line segment/Plane/Perpendicular bisector/Definition
# Triangle/Perpendicular bisector/Circumcenter/Definition

The intersecting point of the three perpendicular bisectors in a nondegenerate triangle in the Euclidean plane is called the circumcenter.
# Linear algebra (Osnabrück 2024-2025)/Perpendicular bisector/Distance condition/Fact/Factreferencenumber

Lemma 37.5
# MDLD/angle bisector

Start=angle bisector Target=Rays/Angle bisector/Definition
# Euclidean space/Distance to subspace/Inner product/Fact

Let $V$ be a Euclidean vector space, $U\subseteq V$ a linear subspace, and $v\in V$. Let $u_{1},\ldots ,u_{m}$ denote an orthonormal basis of $U$.
    $d(v,U)^{2}=\Vert {v}\Vert ^{2}-\sum _{i=1}^{m}\left\langle v,u_{i}\right\rangle ^{2}\,.$
# Linear algebra (Osnabrück 2024-2025)/Euclidean space/Distance to subspace/Inner product/Fact/Factreferencenumber

Corollary 35.7
# MDLD/angle bisectors

Start=angle bisectors See=angle bisector MDLD/angle bisector Target=Rays/Angle bisector/Definition
# MDLD/centroid (triangle)

Start=centroid (triangle) Target=Finite point set/Center of gravity/Definition
# Triangle/Nine-point circle/Fact

Let a nondegenerate triangle $\triangle$ in the Euclidean plane with vertices $A,B,C$ be given. Let $F$ be the circumcircle of the midpoints of the sides of the triangle. Then the following statements hold.
* The radius of $F$ is half of the radius of the circumcircle of $\triangle$.
* The line segements between the orthocenter and the vertices are cut in halves by $F$.
* The feet of the altitudes of $\triangle$ lie on $F$.
# MDLD/orthocenter (triangle)

Start=orthocenter (triangle) Target=Triangle/Orthocenter/Definition
# MDLD/inner product

Start=inner product Target=Vector space/K/Inner product/Definition
# MDLD/inner product (C)

Start=inner product (C) Target=Vector space/K/Inner product/Definition
# MDLD/norm

Start=norm Target=Vector space/K/Norm/Definition
# MDLD/normed vector space

Start=normed vector space Target=Normed vector space/Definition
# MDLD/corresponding metric (norm)

Start=corresponding metric (norm) Target=Normed vector space/Metric/Definition
# MDLD/metric space

Start=metric space Target=Metric/Metric space/Definition
# MDLD/norm (inner product)

Start=norm (inner product) Target=Inner product/K/Norm/Definition
# MDLD/orthogonal vectors

Start=orthogonal vectors Target=Vector space/K/Inner product/Orthogonal/Definition
# MDLD/orthogonal

Start=orthogonal Target=Vector space/K/Inner product/Orthogonal/Definition
# Vector space/K/Inner product/Finite dimension/Linear subspace/Orthogonal projection/Orthonormal basis/Fact

Let $V$ be a finite-dimensional ${\mathbb {K} }$-vector space, endowed with an inner product, and let $U\subseteq V$ denote a linear subspace with an orthonormal basis $u_{1},\ldots ,u_{m}$ of $U$.
orthogonal projection onto $U$ is given by
    $p_{U}(v)=\sum _{i=1}^{m}\left\langle v,u_{i}\right\rangle u_{i}\,.$
# MDLD/orthogonal complement

Start=orthogonal complement Target=Vector space with inner product/Orthogonal complement/Definition
# R^n/Orthogonal complement/Example

Let $V=\mathbb {R} ^{n}$, endowed with the standard inner product. For the one-dimensional linear subspace $\mathbb {R} e_{i}$, generated by the standard vector $e_{i}$, the orthogonal complement consists of all vectors ${\begin{pmatrix}x_{1}\\\vdots \\x_{i-1}\\0\\x_{i+1}\\\vdots \\x_{n}\end{pmatrix}}$, where the $i$-th entry is $0$. For a one-dimensional linear subspace $\mathbb {R} v$, generated by a vector
    $v={\begin{pmatrix}a_{1}\\a_{2}\\\vdots \\a_{n}\end{pmatrix}}\neq 0\,,$
the orthogonal complement can be found by determining the solution space of the linear equation
    $a_{1}x_{1}+\cdots +a_{n}x_{n}=0\,.$
The orthogonal space
    $U=(\mathbb {R} v)^{\perp }={\left\{{\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}\mid a_{1}x_{1}+\cdots +a_{n}x_{n}=0\right\}}\,$
has dimension $n-1$; this is a so-called hyperplane. The vector $v$ is called a normal vector for the hyperplane $U$.
For a linear subspace $U\subseteq \mathbb {R} ^{n}$ that is given by a basis (or a generating system) $v_{i}={\begin{pmatrix}a_{i1}\\\vdots \\a_{in}\end{pmatrix}}$, $i=1,\ldots ,k$, the orthogonal complement is the solution space of the system of linear equations
    $A{\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}=0\,,$
where $A={\left(a_{ij}\right)}$ is the matrix formed by the vectors $v_{i}$ as rows.
# MDLD/orthogonal basis

Start=orthogonal basis Target=Vector space/K/Inner product/Orthogonal basis/Definition
# MDLD/orthonormal basis

Start=orthonormal basis Target=Vector space/K/Inner product/Orthonormal basis/Definition
# MDLD/inner product (K)

Start=inner product (K) Target=Vector space/K/Inner product/Definition
# MDLD/cross product

Start=cross product Target=K^3/Cross product/Direct/Definition
# MDLD/inner products

Start=inner products See=inner product MDLD/inner product Target=Vector space/K/Inner product/Definition
# MDLD/isometry

Start=isometry Target=Vector spaces/K/Inner product/Linear isometry/Definition
# Isometry/Several characterizations with orthonormal basis/Fact

Let $V$ and $W$ be Euclidean vector spaces, and let
    $\varphi \colon V\longrightarrow W$
denote a linear mapping. Then the following statements are equivalent.
* $\varphi$ is an isometry.
* For every orthonormal basis $u_{i}$, $i=1,\ldots ,n$, of $V$, $\varphi (u_{i})$, $i=1,\ldots ,n$, is part of an orthonormal basis of $W$.
* There exists an orthonormal basis $u_{i}$, $i=1,\ldots ,n$, of $V$ such that $\varphi (u_{i})$, $i=1,\ldots ,n$, is part of an orthonormal basis of $W$.
# Euclidean vector space/Isometry with standard space/Fact

For every Euclidean vector space $V$,
isometry
    $\varphi \colon \mathbb {R} ^{n}\longrightarrow V,$
where $\mathbb {R} ^{n}$ carries the
# MDLD/linear isometry (K)

Start=linear isometry (K) Target=Vector spaces/K/Inner product/Linear isometry/Definition
# Linear isometry/Determinant is 1 or -1/Fact

The determinant of a linear isometry
    $\varphi \colon V\longrightarrow V$
on a Euclidean vector space $V$
$1$
$-1$
# Euclidean vector space/Proper isometry/Definition

An isometry on a Euclidean vector space is called proper if its determinant is $1$.
# MDLD/linear isometry

Start=linear isometry Target=Vector spaces/K/Inner product/Linear isometry/Definition
# MDLD/Euclidean vector spaces

Start=Euclidean vector spaces See=Euclidean vector space MDLD/Euclidean vector space Target=Euclidean vector space/Definition
# MDLD/isometry (C)

Start=isometry (C) Target=Vector spaces/K/Inner product/Linear isometry/Definition
# Inner product/Cauchy-Schwarz/Angle/Remark

For vectors $v$ and $w$, different from $0$, in a Euclidean vector space $V$, the inequality of Cauchy-Schwarz implies that
    $-1\leq {\frac {\left\langle v,w\right\rangle }{\Vert {v}\Vert \cdot \Vert {w}\Vert }}\leq 1\,$
holds. Using the trigonometric function cosine (as a bijective mapping $[0,\pi ]\rightarrow [-1,1]$) and its inverse function, the angle between the two vectors can be defined, by setting
    $\angle (v,w):=\arccos {\frac {\left\langle v,w\right\rangle }{\Vert {v}\Vert \cdot \Vert {w}\Vert }}\,.$
The angle is a real number between $0$ and $\pi$. The equation above can be read as
    $\left\langle v,w\right\rangle =\Vert {v}\Vert \cdot \Vert {w}\Vert \cdot \cos {\left(\angle (v,w)\right)}\,.$
This provides the possibility to define the inner product in this way. However, then we have to find an independent definition for the angle. This approach might look a bit more intuitive but is has computationally and in terms of the proofs many disadvantages.
# MDLD/linear isometry (R)

Start=linear isometry (R) Target=Vector spaces/K/Inner product/Linear isometry/Definition
# Euclidean space/Isometry/Structure/Fact

Let
    $\varphi \colon V\longrightarrow V$
be an isometry on the Euclidean vector space $V$.
$V$
orthogonal direct sum
    $V=G_{1}\oplus \cdots \oplus G_{p}\oplus H_{1}\oplus \cdots \oplus H_{q}\oplus E_{1}\oplus \cdots \oplus E_{r}\,$
of $\varphi$-invariant linear subspaces,
$G_{i},H_{j}$
$E_{k}$
$\varphi$
$G_{i}$
$H_{j}$
$E_{k}$
# Linear mapping/Angle-preserving/Definition

A linear mapping
    $\varphi \colon V\longrightarrow W$
between Euclidean vector spaces $V$ and $W$ is called angle-preserving, if for any two vectors $u,v\in V$, the relation
    $\angle (\varphi (u),\varphi (v))=\angle (u,v)\,$
holds.
# MDLD/homotheties

Start=homotheties See=homothety MDLD/homothety Target=Endomorphism/Homothety/Definition
# MDLD/angle-preserving mapping

Start=angle-preserving mapping Target=Linear mapping/Angle-preserving/Definition
# MDLD/angle-preserving (linear)

Start=angle-preserving (linear) Target=Linear mapping/Angle-preserving/Definition
# Euclidean vector space/Linear mapping/Angle-preserving/Structure/Fact

Let
    $\varphi \colon V\longrightarrow V$
denote an angle-preserving linear mapping on the Euclidean vector space $V$.
isometry
    $\psi \colon V\longrightarrow V$
and a homothety
    $\sigma \colon V\longrightarrow V$
such that
    $\varphi =\sigma \circ \psi \,.$
# Linear mapping/Angle-preserving/All angles/Remark

For an angle-preserving mapping
    $\varphi \colon V\longrightarrow W$
between Euclidean vector spaces $V$ and $W$, not only the angles at the origin, but all angles are preserved. For points $P,Q,R\in V$, the angle of the triangle at $Q$ coincides with the angle at $\varphi (Q)$ of the image triangle $\varphi (P),\varphi (Q),\varphi (R)$, because of
    $\angle (P,Q,R)=\angle ({\overrightarrow {QP}},{\overrightarrow {QR}})=\angle (\varphi ({\overrightarrow {QP}}),\varphi ({\overrightarrow {QR}}))=\angle ({\overrightarrow {\varphi (Q)\varphi (P)}}),{\overrightarrow {\varphi (Q)\varphi (R)}})=\angle (\varphi (P),\varphi (Q),\varphi (R))\,.$
# Euclidean space/Distance to subspace/Orthogonal projection/Fact

Let $V$ be a Euclidean vector space, $U\subseteq V$ a linear subspace, and $v\in V$.
$p_{U}(v)$
$U$
$U$
$v$
In particular, we have
    $d(v,U)=d(v,p_{U}(v))\,.$
# Linear algebra (Osnabrück 2024-2025)/Euclidean space/Distance to subspace/Orthogonal projection/Fact/Factreferencenumber

Lemma 35.6
# Linear algebra (Osnabrück 2024-2025)/Vector space/K/Inner product/Finite dimension/Linear subspace/Orthogonal projection/Orthonormal basis/Fact/Factreferencenumber

Lemma 32.14
# MDLD/distance (set)

Start=distance (set) Target=Metric space/Distance/Subsets/Infimum/Definition
# Euclidean space/Affine subspace/Perpendicular/Fact

Let $V$ be a Euclidean vector space, and let $E_{1}=P_{1}+U_{1}$ and $E_{2}=P_{2}+U_{2}$ denote nonempty affine subspaces with the linear subspaces $U_{1},U_{2}\subseteq V$. Let
    $P_{1}-P_{2}=u_{1}+u_{2}+u\,$
with $u_{1}\in U_{1}$, $u_{2}\in U_{2}$, and $u\in {\left(U_{1}+U_{2}\right)}^{\perp }$.
distance $d(E_{1},E_{2})$ equals $\Vert {u}\Vert$; it is obtained in the points $P_{1}-u_{1}\in E_{1}$ and
$P_{2}+u_{2}\in E_{2}$
In particular, the connecting vector of the points, where the minimal distance is obtained, is perpendicular to $E$ and to $F$.
# Distance/Point to affine subspace/Example

Let $E$ be a real affine space over the Euclidean vector space $V$, let $P\in E$ be a point, and let $F\subseteq E$ denote an affine subspace. In case $P\in F$, the distance of $P$ to $F$ equals $0$. In general, we write
    $F=Q+U\,$
with a point $Q\in F$ and with a linear subspace $U\subseteq V$. We determine the orthogonal complement $W=U^{\perp }$ of $U$ in $V$. If $u_{1},\ldots ,u_{m}$ is a basis of $U$, and $w_{1},\ldots ,w_{k}$ is a basis of $W$, then there exists a unique representation
    ${\overrightarrow {PQ}}=\sum _{i=1}^{m}a_{i}u_{i}+\sum _{j=1}^{k}b_{j}w_{j}\,.$
In this case,
    $L=P+\sum _{j=1}^{k}b_{j}w_{j}=Q-\sum _{i=1}^{m}a_{i}u_{i}\,$
is the dropped perpendicular foot of $P$ on $F$, and the distance of $P$ to $L$ is
    $d{\left(P,F\right)}=d{\left(P,L\right)}=\Vert {\sum _{j=1}^{k}b_{j}w_{j}}\Vert \,.$
If the $w_{j}$ form an orthonormal basis of $U$, then this equals ${\sqrt {\sum _{j=1}^{k}b_{j}^{2}}}$.
# Euclidean space/Distance to hyperplane/Linear form/Fact

Let $a\in \mathbb {R} ^{n}$ be a vector with $\Vert {a}\Vert =1$, and let
    $U={\left\{x\in \mathbb {R} ^{n}\mid a_{1}x_{1}+\cdots +a_{n}x_{n}=0\right\}}=(\mathbb {R} a)^{\perp }\,$
denote the linear subspace defined by $a$ as normal vector.
$v\in \mathbb {R} ^{n}$, the distance to $U$ equals
    $d(U,v)=\vert {\left\langle a,v\right\rangle }\vert \,.$