The field of Euclidean triangle geometry has made many advances in the last 150 years or so, and many of the results of from that field are surprisingly accessible with no more than a thorough knowledge of high school geometry. As with many fields, the longer it's around, the more such accessible proofs become available. This blog is, in fact, largely devoted to accessible proofs of seemingly more advanced results.
It is widespread common knowledge in the field of triangle geometry that the incenter of a triangle is also the Nagel point of its medial triangle. The reference cited on the Wikipedia page that mentions this fact merely states it (as an exercise) without proof. As of this writing, searches for proofs online turn up very little. There is a proof in a PDF file on this page (look near the bottom of the page), but that proof uses trigonometry, and is, I've discovered, unnecessarily complicated. In this post, I will retain the structure of that proof, but I will replace the trigonometric portion with a simpler synthetic proof.
Thus, to my knowledge, this is the first synthetic proof of the theorem available on the web.
We begin with a straightforward geometry problem (posed on the website above). Given parallelogram ABCD below, M and N are located such that AM=CN. Prove that DQ bisects angle D of the parallelogram.
The proof of this fact is what the website above completes using trigonometry. I will prove it using the well-known angle bisector theorem. First we extend CM and DA to intersect at point Z as shown below.
Since BC and DZ are parallel, angles Z and ZCB are congruent. The vertical angles at Q therefore make triangle AZQ similar to triangle NCQ. Furthermore, since AB and DC are parallel, triangle AZM is clearly similar to triangle DZC. The first similarity gives the first equals sign below; the stated condition that AM=NC gives the second, and the second similarity gives the third:
Thus consider point Q in triangle DZC. Q divides CZ in the same ratio as the other two sides of that triangle—precisely the configuration for the angle bisector theorem. Since only one point on segment CZ can have that property, the angle bisector theorem guarantees that DQ bisects angle D of the parallelogram, which is what we wanted to prove.
A nice, simple solution to a moderately difficult problem.
But the paper cited above then goes on to use this solution to prove the more important fact we're looking for. Their argument is reformulated below.
We first reveal a larger diagram (in red below) that includes the original problem:
Think of triangle ABC as the original triangle, with three parallelograms (ABCD, ABFC, ACBE) built around it. This makes triangle DEF the anticomplementary triangle of ABC. Or, stated equivalently with different vocabulary, ABC is the medial triangle of DEF. Now let M and N mark the points halfway around the perimeter of ABC from C and A, respectively. Then AM=CN (both are the semiperimeter of ABC minus side AC), and the theorem above applies, and thus DQ is an angle bisector of triangle DEF. This proof could, of course, be repeated for any of the three vertices of DEF.
But when M and N are semiperimeter points in triangle ABC, then Q is the Nagel point. And since Q lies on the angle bisectors of DEF, it is the incenter of that triangle. Thus we have the result we hoped for: The incenter of a triangle is also the Nagel point of its medial triangle.