Proving that in neutral geometry a line cannot be wholly contained in a triangle

Question

Below is a list of (some of) the axioms I'm allowed to use (they are just the usual ones, it's just too much work to list them all). The book doesn't provide a definition for what the interior of a triangle is, but here I'm using that it is the smallest convex set that contains its vertices.

Given any line $r$, there exist points on $r$ and points not on $r$.

Given two different points $A$ and $B$ there exists only one line that contains $A$ and $B$.

Given three points in a line, only one of them is between the other two.

Given two points $A$ and $B$ there always exists a point $C$ between $A$ and $B$ and a point $D$ such that $B$ is between $A$ and $D$.

I'm having a hard time proving the statement. I've tried for a while now but keep making mistakes and making a mess of everything. Any help would be appreciated.

EDIT: To clarify, here the definition of a triangle doesn't include degenerate cases.

Answer 1

Motivated by your description of the interior of a triangle as the convex hull of its vertices, we can say that

A point $X$ is in the interior of $\triangle ABC$ if there exists a point $D$ on line $AB$, with $D$ between $A$ and $B$, such that $X$ lies on line $CD$ and $X$ is between $C$ and $D$.

(I think that is exactly what "convex hull" translates to in neutral geometry, but I have stated this definition in a way that sticks to terms used in Hilbert's axioms. To be fully satisfied with this definition, we ought to prove that its symmetric incarnations, taking vertex $A$ or $B$ as the "third" vertex, describe the same set of points. But that's a different story.)

With this definition, the result you want is a consequence of Pasch's axiom.

Suppose that line $\ell$ contains a point $X$ in the interior of $\triangle ABC$. Let the point $D$ on $AB$ be taken as in the definition above. Then $\ell$ meets side $CD$ of $\triangle ACD$, so either:

• $\ell$ meets line segment $AC$ (a side of the original triangle), or
• $\ell$ meets line segment $AD$ (a subsegment of $AB$, a side of the original triangle).

Either way, we conclude that $\ell$ intersects one of the sides of $\triangle ABC$.

From here, we are pretty much done, and we just need to bring this result back to the definition of "interior of a triangle", above. Without loss of generality, suppose $\ell$ intersects side $AB$ of $\triangle ABC$, and does so at a point $Y$. Then by II.2 of Hilbert's axioms, there is a point $Z$ on line $\ell$ such that $Y$ is between $X$ and $Z$. This point ought to be outside the triangle, since it is on the other side of line $AB$ from $X$, and we will show that indeed, $Z$ cannot be an interior point of $\triangle ABC$.

We proceed by contradiction. Suppose that $Z$ is an interior point, by the definition above, and $D'$ is a point on line $AB$, between $A$ and $B$, that witnesses this. Because $Y$ is between $X$ and $Z$, $Y$ lies on line segment $XZ$ of $\ell$. So consider the intersection points of line $AB$ with $\triangle CXZ$.

• $AB$ meets line $XZ$ at $Y$, which is on the line segment $XZ$.
• $AB$ meets line $CX$ at $D$, which is not on the line segment $CX$ (because $X$ is between $C$ and $D$, which means $D$ cannot be between $C$ and $X$).
• $AB$ meets line $CZ$ at $D'$, which is not on the line segment $CZ$ (because $Z$ is between $C$ and $D'$, which means $D'$ cannot be between $C$ and $Z$).

This contradicts Pasch's axiom, so $Z$ is a point of $\ell$ that isn't interior to $\triangle ABC$.

Below is a diagram of what's going on in this problem. Point $Z$ is shown in red because it cannot actually be there, since it's not in the interior of $\triangle ABC$; by placing it like that, we get a contradiction. (This particular placement of $Z$ is counterfactual to having $Y$ be between $X$ and $Z$.)

Answer 2

Here's an idea: Given a line $\ell_1$ and a triangle, you can extend just one of the sides of the triangle into a line $\ell_2$. Everything inside the triangle is on one side of $\ell_2$. If you're on the other side, you're not in the triangle. As a result, all you need to prove is that there are points of $\ell_1$ on both sides of $\ell_2$.

(We know that $\ell_1$ must intersect one of the sides of the triangle as required. Indeed, any two sides of a triangle meet at a vertex $V$. If we extend the sides into lines, they intersect at $V$ which means that they aren't parallel. After all, if $\ell_1$ is parallel to the first line, and the first line is not parallel to the second, then $\ell_1$ is not parallel to the second. Hence $\ell_1$ is parallel to at most one of the sides; we choose the other one to be $\ell_2$.)

That's intended as a way to get started. If you decide you need more guidance, the following paragraph sketches some ideas for the rest of a proof. There may be many ways to do it.

If the two lines intersect, they intersect at a single point $B$ (we're in neutral geometry). Lines have infinitely many points, so find a second point $A$ on line $\ell_1$. If point $A$ is on the "other" side of $\ell_2$, then it's not in the triangle and so $\ell_1$ isn't either. Otherwise, if point $A$ is on the "inside" side of $\ell_2$, use your last axiom to find point $C$ so that $A-B-C$ are all in a row on $\ell_1$. Point $A$ is on the inside, point $B$ is on the boundary, so point $C$ must be on the other side. This shows that $\ell_1$ has a point that isn't on the inside of the triangle.