RFC: I am wondering if my reasoning below works.

Prove: In a plane, if a line intersects a triangle at a point not a vertex, then it must intersect at least one other side of the triangle.

Let us assume a line intersecting one side of a triangle does not intersect any other side of the triangle.

Let there be a (triangle)ABC and a line L that intersects the triangle at point P. Let P be between (segment)AB.

The points A and B separate a plane into two half planes H1 and H2. From the definition of a triangle, one point is non-collinear with the other two points that make the triangle. Thus, point C will lie in either H1 or H2.

Since A-P-B (P is between A and B) with P on L, there is at least one point of L in the interior of (triangle)ABC.

From the definition of a triangle, the perimeter of a triangle is the sum of the length of its sides. Since a perimeter can not be infinite it must be finite and subsequently the length of the sides must be finite.

A line has no finite length by definition as that would be a segment. Therefore line L may have a segment from point P longer than the sum of the legs of the triangle. Placing this segment on the same half plane as point C of the triangle, the endpoint opposite P will not lie in the interior of the triangle ABC.

With one side of the segment lying in the interior of triangle ABC and the other outside the interior we contradict our assertion that the line L does not intersect another leg of the triangle ABC.

Therefore, a line through one side of a triangle must intersect at least one other leg.


2 Answers 2


I think your reasoning is correct, but it's a bit long and not polish. I would suggest this simplified argument based on your idea:

Let $P$ be in the segment $AB$ but distinct from $A$ and $B$. Assume a line $L$ passing through $P$ divides the plane in $2$ partitions, with $A$ and $B$ lying on different ones.

But the perimeter of a triangle is a closed path, and in particular part of it goes from $A$ to $B$ staying away from $AB$. To do so it needs to pass from one partition to the other, and hence it necessarily passes through $L$ again, the intersection point being different from $P$.

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    $\begingroup$ Delightful. Thank you. $\endgroup$ Dec 3, 2013 at 3:26
  • $\begingroup$ How about, If $C$ is on $L$, then $L$ intersects $\overline{AC}$ and $\overline{BC}$.Otherwise, $C$ and $A$ are on opposite sides of $L$ or $C$ and $B$ are on opposite sides of $L$.. . . $\endgroup$ May 15, 2015 at 19:41

See Pasch's axiom. Basically, most contemporary formal treatments assume that what you are trying to prove is an axiom. If you want it to be a theorem, you'll need some other "plane order" axiom.


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