Geometry

# Triangle Centers: Circumcenter

Moving forward from perpendicular bisectors, let’s talk about a topic you’ll certainly see in a geometry course: triangle centers.

If you Google “triangle centers“, you’ll find that there are actually compiled indexes of these creatures – they number in the hundreds at least. In fact, there are so many that they have to be named using alphanumeric codes! To keep this site from turning into another exhaustive list, we will cover the four triangle centers that are by far the most common ones you will see.

### Part 1: Circumcenter

Hint: We’ll need our knowledge from the last post on perpendicular bisectors to figure out this triangle center.

The circumcenter is the point on the triangle that is equidistant from each of the triangle’s vertices. To put it another way, the lines connecting the circumcenter and each vertex are the same length. So we can say that the circumcenter is the center of a circle (aptly named the circumcircle) that passes through the triangle’s vertices. In the figure below, D is the circumcenter and the center of the circumcircle.

We’re given a triangle, like this:

How does one construct a circumcenter? Remember, we’re looking for a point that is equidistant from the triangle vertices. As it happens, any point equidistant from the vertices of a line is on the perpendicular bisector of that line. (Recall the arcs drawn in the last post.) So let’s find the perpendicular bisectors of two of the triangle sides (three, if you want, but you only need two lines to find an intersection).

In this diagram, we have constructed the perpendicular bisectors of AB and AC. Any point on the bisector of AB is equidistant from A and B, and any point on AC’s bisector is equidistant from A and C. Logically, if we take the intersection of the two bisectors D, then D is equidistant from A, B, and C. Therefore, D is the circumcenter of triangle ABC.

To summarize: The circumcenter of a triangle is the intersection of two of the triangle’s perpendicular bisectors because the intersection is equidistant from each of the vertices of the triangle.