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# Pasch’s theorem

###### Theorem.

(Pasch) Let $\triangle abc$ be a triangle with non-collinear vertices $a,b,c$ in a linear ordered geometry. Suppose a line $\ell$ intersects one side, say open line segment $\overline{ab}$, at a point strictly between $a$ and $b$, then $\ell$ also intersects exactly one of the following:

$\overline{bc}\mbox{, }\qquad\qquad\overline{ac}\mbox{, }\qquad\qquad c.$ |

###### Proof.

First, note that vertices $a$ and $b$ are on opposite sides of line $\ell$. Then either $c$ lies on $\ell$, or $c$ does not. if $c$ does not, then it must lie on either side (half plane) of $\ell$. In other words, $c$ and $a$ must be on the opposite sides of $\ell$, or $c$ and $b$ must be on the opposite sides of $\ell$. If $c$ and $a$ are on the opposite sides, $\ell$ has a non-empty intersection with $\overline{ac}$. But if $c$ and $a$ are on the opposite sides, then $c$ and $b$ are on the same side, which means that $\overline{bc}$ does not intersect $\ell$. ∎

Remark A companion property states that if line $\ell$ passes through one vertex $a$ of a triangle $\triangle abc$ and at least one other point on $\triangle abc$, then it must intersect exactly one of the following:

$b\mbox{, }\qquad\qquad c\mbox{, }\qquad\qquad\overline{bc}.$ |

Of course, if $\ell$ passes through $b$, $\overline{ab}$ must lie on $\ell$. Similarly, $\overline{ac}$ lies on $\ell$ if $\ell$ passes through $c$.

## Mathematics Subject Classification

51G05*no label found*

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