Circumcenters and Perpendicular Bisectors

($\Rightarrow$) Suppose $P$ lies on the perpendicular bisector of $AB$, meeting $AB$ at $M$. Then $AM=MB$, $PM=PM$, and $\angle PMA=\angle PMB=90^\circ$.

so $PA=PB$.

($\Leftarrow$) Suppose $PA=PB$. Let $M$ be the midpoint of $AB$. Since $AM=MB$ and $PM=PM$,

Thus $\angle PMA=\angle PMB$. These angles are supplementary, so both equal $90^\circ$. Hence $PM\perp AB$, and $P$ lies on the perpendicular bisector.

Concurrency

Draw the perpendicular bisectors of $BC$ and $AC$, and let them intersect at $O$. We must prove that $O$ also lies on the perpendicular bisector of $AB$.

Since $O$ lies on the perpendicular bisector of $BC$,

Since $O$ lies on the perpendicular bisector of $AC$,

Therefore,

By the converse of the perpendicular bisector theorem, $O$ lies on the perpendicular bisector of $AB$. Thus all three perpendicular bisectors are concurrent at $O$, and

For an acute triangle:

By the inscribed angle theorem, the minor arc $BC$ has measure $2A$ when $A<90^\circ$. Hence

In isosceles triangle $BOC$, $OB=OC$, so

Similarly, the minor arcs $CA$ and $AB$ have measures $2B$ and $2C$, respectively, when the corresponding angles are acute.

For an obtuse triangle, suppose $A>90^\circ$. The arc $BDC$ not containing $A$ is the major arc $BC$, so

Therefore the reflex angle $BOC$ is $2A$, while the ordinary smaller angle is

Similarly, because $B,C<90^\circ$,

In isosceles triangle $BOC$,

In isosceles triangles $AOC$ and $AOB$,

For the illustrated acute triangle, $OM\perp BC$ and $BM=MC=a/2$. Since $\angle BOC=2A$, we have $\angle BOM=\angle MOC=A$.

In right triangle $BOM$,

Thus

Similarly,

Since $[ABC]=\frac12bc\sin A$, we obtain

In right triangle $BOM$,

for acute $A$. Also, by the Pythagorean theorem,

For obtuse $A$, $\cos A<0$ but $OM$ is a nonnegative distance. Therefore the general formula is

In triangle $ABC$, prove that $AH=2OM$ if $ABC$ is acute, where $H$ is the orthocenter, $O$ the circumcenter, and $M$ the midpoint of $BC$.

A triangle has side lengths $13$, $14$, and $15$. Find its circumradius and the distance from its circumcenter to the side of length $14$.

Let $ABC$ be a triangle with circumcenter $O$. Suppose

Find the circumradius, the distance from $O$ to $BC$, and $\angle BOC$.

The circumcenter is the midpoint of the hypotenuse.

If $\angle A=90^\circ$, then

In particular, the midpoint of the hypotenuse is equidistant from all three vertices.

For an equilateral triangle of side length $a$, the circumcenter, incenter, centroid, and orthocenter coincide.

The circumradius is

The circumcenter lies outside the triangle, on the opposite side of the longest side from the obtuse vertex.

If $A>90^\circ$, then

where $M$ is the midpoint of $BC$.

For

the circumcenter $O=(u,v)$ satisfies

Subtracting the squared-distance equations gives a linear system:

The circumcenter is inside an acute triangle, at the midpoint of the hypotenuse of a right triangle, and outside an obtuse triangle.

Incorrect as a general formula:

Correct for every nondegenerate triangle:

The relation

does not always hold for the ordinary, smaller central angle.

If $A>90^\circ$, then

Three noncollinear points determine exactly one circle.

If three distinct points are collinear, no circle passes through all three.