Prove a line cannot intersect all 3 of the sides of a triangle at interior points

Here, a story is told of a competitor in the Russian math olympiad of 1945. The boy did not solve a single problem, but received a prize for writing:

I spent much time trying to prove that a straight line can't intersect three sides of a triangle in their interior points but failed for, to my consternation, I realized that I have no notion of what a straight line is.

Assume the standard Euclidean geometry of the plane. What is the simplest way to prove this statement? Can it be done without analytic geometry?

As noted in the quote, a lot of this has to do with how to rigorously define. A related question is: how is Euclidean geometry rigorously defined today, and in particular, what is a line? What is a point? A triangle?

Yunone is right that this fact is basically equivalent to Pasch's postulate, which can't be proved from Euclid's axioms. There are many such gaps in the axioms corresponding to incidence results that are obviously true if you draw a picture and hence were left unstated, like that the two circles in his first proposition intersect, and that is why Euclid's axioms are said to be insufficient.

So, one answer to your question is that the result in question is true as a consequence of Pasch's postulate if you are using an axiom system that includes it. But there is no geometric axiom system that can be considered the standard formulation of Euclidean geometry today. Rather, the standard formulation nowadays is in terms of the Cartesian plane. Points are ordered pairs, lines are solution sets of linear equations, distance is given by the usual formula, etc.

In this formulation Pasch's postulate is a theorem that can be proven.

A straightforward way to see it is as a consequence of Menelaus' theorem, which can be proven purely algebraically. If a line intersected the interiors of the three sides, this theorem would give us the product of three positive numbers as -1.

But there's a more direct though ugly way to see it. Given a line intersecting the interiors of all sides of a triangle, you can use coordinates to transform this statement into a series of equations and inequalities between real numbers and show they cannot be satisfied. Here are the unattractive details:

Translations preserve lines and betweenness, so translate one vertex to the origin. Linear transformations also preserve lines and betweennes, so transform the two other vertices to (0,1) and (1,0). So, without loss of generality we can assume our triangle has vertices (0,0), (0,1), and (1,0). Now, suppose a line intersects all three sides. If it were vertical, it would be parallel to one side and thus couldn't intersect all three. Thus, the line is the graph of a function f(x). Suppose it intersects the interior of the two edges meeting the origin. Then

$0 < f(0) < 1$ and there is an r in $(0,1)$ such that $f(r) = 0$. Thus $f(x)$ is decreasing so it can be written as $f(x) = mx + b$ where$ m < 0$ and $0 < b < 1$, and also $f(1)=m+b<f(r)=0$ from the fact it is decreasing.

$1-x$ is the equation of the third side. Let's look at the intersection between this side and the line defined by $f(x)$. If $1-x = mx+b$, then $(m+1)x=1-b$. If it intersects the interior of this side, $0 < x < 1$. Thus $m+1 > 0$, so $1-b<m+1$, so $m+b <0$. But $m+b > 0$ was proved above. So it's impossible for any line to intersect the interiors of all three sides.

This proof is ugly, but it demonstrates that Pasch's postulate is a theorem that you can prove in the Cartesian model of Euclidean geometry.

A line l divides the plane into two halves. Out of three points of the triangle A, B, C, two of them will definitely lie on one half (by the Pigeonhole principle and noting that none of them lie on the line itself because of the initial problem condition). Let these two points be A, B. Then the line l does not intersect the line AB of the triangle.

In a rather hand-waving way and assuming (a) the straight line cannot intersect a side more than once, (b) intersecting a side at an interior point means passing from inside to outside the triangle or from outside to inside, and (c) the straight line starts and ends outside the triangle:

A straight line starts outside the triangle. If it intersects a side of the triangle at an interior point, it passes inside the triangle. If it intersects another side of the triangle at an interior point, it passes outside the triangle. If it intersects a third side of the triangle at an interior point, it passes inside the triangle, but that is impossible as it ends outside the triangle and there are no more sides to intersect.