Suppose there is a straight line segment \(\overline{\small\mathtt{AB}}\) between you, at point \(\small\mathtt{A}\), and some point \(\small \mathtt{B}\) you want to reach. Is there a shorter way to point \(\small\mathtt{B}\) that uses 2 line segment paths?
If you think the answer is yes, or if you think that the answer is no but it's not obviously no, the Epicureans have some pretty mean things to say about you:
The theorem mentioned is the Triangle Inequality Theorem: the sum of the lengths of any two sides of a triangle is always greater than the length of the remaining side. And it is fascinating to me how seldom I have seen this theorem in textbooks phrased as a "shortest straight-line distance" statement. Most often, I see investigations about what side lengths can make up a triangle, which turns the ridiculously obvious into a complicated issue. But let's discuss that below. For now, a proof!
It's Okay To Be Both Obvious and Require Proof
We want to show that \(\small\mathtt{BA + AC}\) is greater than \(\small\mathtt{BC}\), that \(\small\mathtt{AB + BC}\) is greater than \(\small \mathtt{AC}\), and that \(\small\mathtt{BC + CA}\) is greater than \(\small\mathtt{AB}\).
Euclid starts by extending \(\overline{\small\mathtt{BA}}\) to a point \(\small\mathtt{D}\) such that \(\small\mathtt{DA = AC}\) as I've shown in the diagram. This means that \(\small\Delta\mathtt{ADC}\) is an isosceles triangle, and \(\small\measuredangle\mathtt{ACD}\) and \(\small \measuredangle\mathtt{ADC}\) are congruent. And since "the whole is greater than the part," m\(\small\measuredangle\mathtt{BCD}\) is greater than m\(\small\measuredangle\mathtt{ACD}\), which also makes m\(\small\measuredangle\mathtt{BCD}\) greater than m\(\small\measuredangle\mathtt{ADC}\).
Dizzy yet? Just remember the last rung of the ladder we got to: m\(\small\measuredangle\mathtt{BCD}\) is greater than m\(\small\measuredangle \mathtt{ADC}\). Therefore, we can say something about the line segments that those two angles "catch." Specifically, we can say that \(\overline {\small\mathtt{DB}}\) is longer than \(\overline{\small\mathtt{BC}}\), because Proposition 19. This is the same as saying that \(\small\mathtt{DA + AB > BC}\). And since \(\small\mathtt{DA = AC}\), we can make a substitution to get \(\small \mathtt{AC + AB > BC}\), which is the first of the three statements above that we wanted to prove: \(\small\mathtt{BA + AC}\) is greater than \(\small \mathtt{BC}\). Euclid tells us that we can prove the other two statements with a similar method, and I believe him.
Internalizing the Idea That Mathematics Is Complex and Intuition-Free
In the link is an example of what we often put students through to investigate the Triangle Inequality Theorem. This is consistent with what I have seen in a lot of lesson plans.
Were the authors of this document aware that there is a very intuitive way of looking at this theorem? It might be enough to suppose that they weren't and that they, like all of us, sometimes just repeat what we see or hear elsewhere.
But it's also worth entertaining the possibility that they did know and pressed on anyway. What good reasons might they have for doing so? I think the best answer is simply Proclus's reply above: "a mere perception of the truth of the theorem is a different thing from a scientific proof of it and a knowledge of the reason why it is true." To which I would respond, Indeed, a different thing. Not necessarily a better thing.
If we can give students the "mere perception of the truth" of a theorem (or of any mathematical idea), we should do so, even if it doesn't make us feel smart, leaves a lot of class time to fill, or runs counter to a set of standards. I would argue that students still have to prove those theorems and justify those ideas. But they can then do so correctly oriented to the reality of what they are doing: drawing on their own perceptions and knowledge to make their ideas plain to others.