Circumcentre - concurrency of the perpendicular bisectors

The circumcentre is the intersection of the three perpendicular bisectors of the sides and it is equidistant from the three vertices: that is why it is the centre of the circumscribed circle.

Theorem — Concurrency of the perpendicular bisectors

The three perpendicular bisectors of the sides of a triangle meet at a single point $O$ (circumcentre), equidistant from the three vertices: $OA\cong OB\cong OC$.

Proof — The perpendicular bisectors are concurrent

1. Let $m_1$ be the perpendicular bisector of $AB$ and $m_2$ that of $BC$, and let $O$ be their point of intersection.
2. Since $O$ lies on the perpendicular bisector of $AB$, by the locus theorem we have $OA\cong OB$.
3. Since $O$ lies on the perpendicular bisector of $BC$, we have $OB\cong OC$.
4. By transitivity: $OA\cong OB\cong OC$, hence $O$ is equidistant from the three vertices.
5. But then $OA\cong OC$ implies that $O$ also lies on the perpendicular bisector of $AC$ (the third one): the three perpendicular bisectors are concurrent at $O$, the centre of the circumscribed circle.

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Remark — Circumcentre in the right triangle

In a right triangle the circumcentre coincides with the midpoint of the hypotenuse.

Indeed the angle at $C$ is right ($= \frac{\pi}{2}$) and is an inscribed angle subtending the arc $AB$. The corresponding central angle is $\pi$, hence $A$, $O$ and $B$ are collinear: $O$ lies on the hypotenuse. Since $OA\cong OB$ (radii), $O$ is the midpoint.

Links

Topics: Euclidean circle
Concepts: Perpendicular bisector · Circumcentre · Circumscribed circle
Skills: Proving · Synthetic geometry