(by Polymath) This page is part of the Advanced High School Math Project.
While good proofs of the existence of the four classical centers (those known to the ancients: the circumcenter, the incenter, the orthocenter and the centroid), are readily available, I will give some proofs here for completeness.
Triangle centers are tricky to define technically, but (roughly) they are points of concurrence of cevians (or other important lines in a triangle). If this concept is unfamiliar to you, or if you don't know what Ceva's theorem says, you should read this post first, because we will use the theorem several times here.
We begin with the concurrence of the perpendicular bisectors of the sides of a triangle. Below is a picture showing those three lines. They do appear in this diagram to concur; however, that's what we're trying to prove, so I will be careful not to assume (based on that appearance) that they do in fact concur.
The red, green, and heavy black lines are the perpendicular bisectors of the sides of the triangle, and define point Q to be the point where the red and green ones intersect. We are out to prove that Q also lies on the black one, since that would mean that the lines concur. The way to see this is to remember that a perpendicular bisector of a segment is precisely the set of all points in the plane that are equidistant from the endpoints of that segment. Thus any point on the red line (which perpendicularly bisects BC) must be equidistant from B and C. Thus QB and QC are congruent. For the same reason, except using the green line, QA and QC are congruent. Which obviously means that QA and QB are congruent. But then Q is equidistant from A and B, and thus falls on AB's perpendicular bisector: the black one, which is just exactly what we wanted to prove.
Thus Q falls on all 3 perpendicular bisectors, and it must be equidistant from each of the vertices. Hmmmmmm....a single point equidistant from three others? That sounds like...there's a circle hiding here! Indeed, it means there's a circle centered at Q that goes through the three vertices of the triangle. Since the circle circumscribes the triangle, it's called the circumcircle, and Q is called the circumcenter of the triangle.
You can make a very similar argument about the angle bisectors of the triangle, since angle bisectors are precisely the set of all points in the plane that are equidistant from the rays of an angle. (Angle bisectors and perpendicular bisectors are duals of each other.) But in order to show the power of Ceva's theorem, I will prove the rest of the concurrence theorems using it.
Recall that to use Ceva's theorem, we construct a product of ratios and prove that the product equals 1. Let's use this first to prove that the medians of a triangle concur (drawn below, again appearing to concur despite not having proved it yet).
Since the three cevians are medians, the ratios of the parts of each side of the triangle (AZ:ZB, for instance) are all obviously 1:1. But those ratios are exactly the ones that appear in Ceva's product:
Since proving that this ratio equals 1 is all we need, we're done already—the medians concur! Their point of concurrence is called the centroid of the triangle, and it has the distinction of being the center of gravity of the triangle; if the triangle were made of cardboard, it would balance on a pinpoint placed at the centroid.
Proving that the angle bisectors concur is almost just as easy if you remember the angle bisector theorem. It says that an angle bisector of a triangle divides the side where it lands into segments that stand in the same ratio as the two remaining, adjacent sides of the triangle. So, for instance, AZ:ZB is the same as b:a (using the standard notation for triangles so that lower case b stands for the length of the side opposite angle B).
Using the theorem on Ceva's product we get:
And thus the angle bisectors concur at Q. But furthermore, noting the earlier discussion of angle bisectors, Q must be equidistant from the three sides of the triangle, and thus it is the center of a circle that just barely touches the sides: the inscribed circle. Hence the concurrence point of the angle bisectors is called the incenter.
Finally, we look at the altitudes of a triangle. The picture below shows the case for an acute triangle, but the proof remains almost identical for an obtuse one. The only difference is that in obtuse triangles the inner similar triangles are similar for other reasons. The resulting ratios, however, stay the same, as does the rest of the proof.
In the first step below, the only change was to rearrange the segments in the denominators; but since the fractions are being multiplied, that doesn't change the numerical value of the product.
In the second step, I've replaced each ratio with another ratio of sides of the triangle. This replacement is based on various pairs of similar triangles: for instance, triangles AZC and AYB are similar (they share angle A and they each have a right angle), and so the ratio of their legs (AZ:AB) is the same as the ratio of their hypotenuses (b:c). Repeat that argument twice for the other ratios, and we get another expression that clearly reduces to 1. Thus the altitudes of any triangle concur, and their point of concurrence is called the orthocenter.
The fact that all these segments in a triangle concur is, at first, kind of amazing. Concurrence turns out to be so common, however, that you will start to take these basic centers for granted. I've heard it said that there are "more miracles per square mile" to discover in the field of triangle geometry than in most other mathematical fields. The availability of geometry exploration software has made it relatively common for amateurs to contribute to the field, though mathematicians have been coming to understand the ideas at deeper and deeper levels recently (you know there must be some significant math going on when John Conway is getting involved).
- The perpendicular bisectors of a triangle concur at the circumcenter, which is the center of the circumcircle (going through all three vertices of the triangle).
- The medians concur at the centroid, which is the center of gravity of the triangle.
- The angle bisectors concur at the incenter, which is the center of the incircle (tangent to all three sides of the triangle).
- The altitudes concur at the orthocenter.