In \(\small\Delta\)\(\small\mathtt{ABC}\), point \(\small\mathtt{D}\) is fixed at the midpoint of \(\overline{\small\mathtt{AC}}\), and point \(\small\mathtt{E}\) is fixed at the midpoint of \(\overline{\small\mathtt{BC}}\). The Triangle Midsegment Theorem boldly says that no matter how you mess with that triangle, \(\overline{\small\mathtt{DE}}\) will be parallel to \(\overline{\small\mathtt{AB}}\) and also half the length of \(\overline{\small\mathtt{AB}}\).
You can click and drag points \(\small\mathtt{A}\), \(\small\mathtt{B}\), and \(\small\mathtt{C}\) to see that the theorem at least seems to be true: \(\overline{\small\mathtt{AB}}\) and \(\overline{\small\mathtt{DE}}\) seem to stay parallel, and \(\overline{\small\mathtt{DE}}\) looks like it stays at about half the length of \(\overline{\small\mathtt{AB}}\).
Side-Angle-Side
Notice that we've got two triangles, \(\small\Delta\mathtt{ABC}\) and \(\small\Delta\mathtt{DEC}\). The two triangles share an angle, \(\small \measuredangle\mathtt{C}\). Plus, since the midpoints divide the two sides into two congruent segments, the ratios of the corresponding side lengths of the big triangle and small triangle are equal: \(\small\mathtt{AC}\) : \(\small\mathtt{DC}\) = \(\small\mathtt{BC}\) : \(\small\mathtt{EC}\) = 2 : 1. This means, by the Side-Angle-Side Similarity Theorem (or is it a postulate?), that all of the corresponding side lengths have the same ratio. So \(\overline{\small \mathtt{AB}}\) has twice the length of \(\overline{\small\mathtt{DE}}\).
But Are They Parallel?
That's one part down. The other part of the theorem mentioned that the two line segments (the midsegment and the "third" side) are always parallel. How can we show that that's true?
Well, since \(\small\Delta\mathtt{ABC}\) and \(\small\Delta\mathtt{DEC}\) are similar, their corresponding angles are congruent. So, \(\small
\measuredangle\mathtt{A}\) \(\small\cong\) \(\small\measuredangle\mathtt{D}\) and \(\small\measuredangle\mathtt{B}\) \(\small\cong\) \(\small
\measuredangle\mathtt{E}\).
This matters because you can think of imaginary lines running through \(\overline{\small\mathtt{DE}}\) and \(\overline{\small\mathtt{AB}}\), which are
"cut" by a transversal--either line \(\small\mathtt{AC}\) or line \(\small
\mathtt{BC}\). And when corresponding angles are congruent in that
situation, the lines must be parallel.
Confusion and Clarification Points
What I really want to do with this theorem is demonstrate it algebraically, using linear functions on the coordinate plane. I've included a Desmos graph below that does this--but with too much trigonometry for my taste.
One reason I'm more tempted to approach this through algebra is that it is not obvious to me that the two triangles remain similar even when I tug on the vertices to distort them. Maybe the better idea to have in my head is that no matter how many different triangles I draw, I can always pick two sides, determine their midpoints, and connect these midpoints to form two similar triangles.
Of course, that raises the question, why \(\small\frac{1}{2}\)? Drawing the line segment anywhere to connect two sides gets me two similar triangles, doesn't it? Or maybe not just anywhere. But drawing the midsegment of the smaller triangle in the example above would create three similar triangles, no? And that tiny midsegment would divide the two sides of the larger triangle into a segment that is \(\small\frac{1}{4}\) the length of the whole and a segment that is \(\small\frac{3}{4}\) the length of the whole.
The parallel part is confusing too, mostly because when we talk about transversals, we're almost always talking about things that cross lines that we already know are parallel. It seems like I didn't spend enough time nailing down the fact that as long as two lines aren't coincident, they have a transversal. I think.