# Circumcenter

The circumcenter is the center of the circumcircle of a polygon. Only certain polygons can be circumscribed by a circle: all nondegenerate triangles have a circumcircle whose circumcenter is the intersection of the perpendicular bisectors of the sides of the triangle. Quadrilaterals which have circumcircles are called cyclic quadrilaterals. Also, every regular polygon is cyclic.

## Proof that the perpendicular bisectors of a triangle are concurrent

### First Proof

We consider a nondegenerate triangle $\triangle ABC$. Since the triangle is nondegenerate, $AB$ and $BC$ lie on different lines and so their perpendicular bisectors are not parallel and thus intersect. Let $O$ be the intersection of these perpendicular bisectors. Since $O$ lies on the perpendicular bisector of $AB$, it is equidistant from $A$ and $B$; likewise, it is equidistant from $B$ and ${C}$. Hence $O$ is equidistant from $A$ and ${C}$; hence $O$ also lies on the perpendicular bisector of $AC$ (and is the circumcenter).

### Second Proof

One of the most common techniques for proving the concurrency of lines is Ceva's Theorem. However, there aren't any cevians in the diagram which would be needed for a direct application of Ceva's Theorem. Thus, we look for a way to make some by drawing in helpful lines. Drawing in $DE, EF$ and $FD$ (i.e. the medial triangle of $ABC$) does the trick.
By SAS Similarity $\triangle BFD\sim \triangle BAC$. Thus $\angle BFD = \angle BAC$ making $FD || AC$. Since $EO\perp AC$ and $AC\| FD, EO\perp FD$ making $EH$ an altitude of $DEF$. Likewise, $DG$ and $FI$ are also altitudes. Thus, the problem is reduced to proving that the altitudes of a triangle are concurrent. This can be done using Ceva's Theorem.