# Intercept theorems/Linear algebra/Section

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.

## Theorem

Let $V$ denote a two-dimensional Euclidean vector space, and let $s_{1},s_{2},v\in V$ be vectors and suppose that $v$ and $s_{1}$ are 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.

## Corollary

Let $V$ denote a two-dimensional Euclidean vector space, and let $s_{1},s_{2},v\in V$ be vectors and suppose that $v$ and $s_{1}$ are 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.

## Theorem
### Proof

$\Box$