by Polymath This page is part of the Advanced High School Math Project.

The nine-point center of a triangle is the center of a circle containing, as we'll see below, nine significant points of the circle. That circle was discovered (more or less, according to this article) by Feuerbach and studied by Euler, so the circle is sometimes called the Feuerbach circle or the Euler circle.

Until I discovered the proof below (in an abbreviated form) in the source cited below, I thought the proof of the existence of the nine-point circle was long and complicated. But this proof is remarkably accessible and reasonably short.

Let's get right to it. First a preparatory lemma. Draw a circle using a side of any triangle as its diameter, and then connect an endpoint of that diameter to the point on the opposite side of the triangle where the circle intersects it. This is shown in the diagram below.

Since angle AHB is inscribed in a semicircle it is a right angle, which means that BH is the altitude from point B. The same argument works from the other side, of course, using the altitude from A. The lemma just says, then, that a circle whose diameter is side AB of a triangle contains the feet of the altitudes from A and B. Note that this works just as well on an obtuse triangle. Simple enough.

Consider next the diagram below.

Segment BH is the altitude we just discussed, and the points M, N, and P are the midpoints of the three sides of the triangle. We're interested in the red quadrilateral HNPM. Since MP is a midline of the triangle, it is parallel to HN, making that quadrilateral a trapezoid.

Since NP is also a midline of the triangle, it is half the length of AB. But now compare the diagram to the one above it. Note that M is obviously the center of the circle (since its diameter was AB), and that makes MH a radius of the circle, and therefore half the length of AB—just like NP. Brilliant! That trapezoid is therefore actually an isosceles trapezoid.

Since angles P and N of the trapezoid are supplementary (due to the parallel lines), and angles M and P are congruent (due to the trapezoid's isosceles-ness) the opposite angles of that trapezoid are supplementary, which makes it a cyclic quadrilateral (why?). So that means that the circle that contains the three midpoints of the sides also contains H. And, of course, we can repeat the proof for all three of the altitude feet, giving a circle that goes through all six points.

That's the circle we're looking for. But I promised nine points, not just six. So consider the next diagram, in which the three altitudes, and thus the orthocenter O are drawn. I'll explain points X and Y below.

Since 3 points completely determine a circle, the three feet of the altitudes of *any* triangle must be on its nine-point circle, and that circle must thus contain the three midpoints of the sides. As we showed, that's true for triangle ABC above, but it's also true for triangle BOC. Look carefully at its altitudes. They're segments JB, CH, and OK. So the three feet are J, K, and H—the very same feet of the altitudes of triangle ABC. Thus the circle we found above is also the nine-point circle for the smaller triangle BOC, and it must pass through the midpoints of *its* sides. Those midpoints are P (we already knew the circle passed through P), X, and Y. And repeating the argument would prove that the circle also contains Z, the midpoint of segment AO.

So. In every triangle, there's a circle passing through nine points: the three midpoints of the sides, the three feet of the altitudes, and the three midpoints of the segments connecting each vertex to the orthocenter. That's the nine-point circle, and its center is the called the nine-point center. It's demonstrated below.

You can read more about it here and here. The only other fact I'll mention in this essay is that its existence proves an "is-also" theorem. The triangle created from the feet of the altitudes is called the orthic triangle, and the one created from the midpoints of the sides is called the medial triangle. The nine-point circle clearly circumscribes them both, so the theorem is "The circumcenter of the orthic triangle is also the circumcenter of the medial triangle. Furthermore, that point is also the nine-point center of the original triangle."

Source: Dorrie