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 lines 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, 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.

An application of the intercept theorem. One can compute the width of the river without crossing the river.

Another application of the intercept theorem is to guess a distance using one thumb but two eyes.

Corollary

Let ${}V$ denote a two-dimensional Euclidean vector space, 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

Let ${}V$ be a two-dimensional Euclidean vector space, let ${}s_{1},s_{2},s_{3}\in V$ be vectors, and suppose that ${}v\in V$ is 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

${}{\frac {d(B_{1},B_{2})}{d(A_{1},A_{2})}}={\frac {\Vert {B_{2}}\Vert }{\Vert {A_{2}}\Vert }}={\frac {d(B_{2},B_{3})}{d(A_{2},A_{3})}}\,.$

$\Box$