8.1: Circumcircle and circumcenter
Perpendicular bisectors to the sides of any nondegener- ate triangle intersect at one point.
The point of intersection of the perpendicular bisectors is called circumcenter. It is the center of the circumcircle of the triangle; that is, a circle that passes thru all three vertices of the triangle. The circumcenter of the triangle is usually denoted by \(O\).
- Proof
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Let \(\triangle ABC\) be nondegenerate. Let \(\ell\) and \(m\) be perpendicular bisectors to sides \([AB]\) and \([AC]\) respectively.
Assume \(\ell\) and \(m\) intersect, let \(O = \ell \cap m\).
Let us apply Theorem 5.2.1 . Since \(O \in \ell\), we have that \(OA = OB\) and since \(O \in m\), we have that \(OA = OC\). It follows that \(OB = OC\); that is, \(O\) lies on the perpendicular bisector to \([BC]\).
It remains to show that \(\ell \nparallel m\); assume the contrary. Since \(\ell \perp (AB)\) and \(m \perp (AC)\), we get that \((AC) \parallel (AB)\) (see Exercise 7.1.1 ). Therefore, by Theorem 5.3.1 , \((AC) = (AB)\); that is, \(\triangle ABC\) is degenerate - a contradiction.
There is a unique circle that passes thru the vertexes of a given nondegenerate triangle in the Euclidean plane.
- Answer
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Apply Theorem \(\PageIndex{1}\) and Theorem 5.2.1 .