Pasch’s theorem

Theorem.

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

$$\overline{bc}\mathit{\text{, }}\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\overline{ac}\mathit{\text{, }}\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}c.$$

Proof.

First, note that vertices $a$ and $b$ are on opposite sides of line $\mathrm{\ell }$. Then either $c$ lies on $\mathrm{\ell }$, or $c$ does not. if $c$ does not, then it must lie on either side (half plane) of $\mathrm{\ell }$. In other words, $c$ and $a$ must be on the opposite sides of $\mathrm{\ell }$, or $c$ and $b$ must be on the opposite sides of $\mathrm{\ell }$.If $c$ and $a$ are on the opposite sides, $\mathrm{\ell }$ 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 $\mathrm{\ell }$.∎

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

$$b\text{, }\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}c\text{, }\mathit{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}\overline{bc}.$$

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