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# 8.1: Circumcircle and circumcenter

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Theorem $$\PageIndex{1}$$

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

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.

Exercise $$\PageIndex{1}$$

There is a unique circle that passes thru the vertexes of a given nondegenerate triangle in the Euclidean plane.

Answer

Apply Theorem $$\PageIndex{1}$$ and Theorem 5.2.1.