Perpendicular Bisectors & the Circumcenter

The perpendicular bisector theorem

The theorem says: if a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. So a point $P$ on the perpendicular bisector of $\overline{AB}$ satisfies $PA = PB$. The converse is also true — if $PA = PB$, then $P$ must lie on the perpendicular bisector of $\overline{AB}$.

In a figure, look for two clues: tick marks showing the segment is cut into two congruent pieces, and a right-angle mark where the bisector crosses. When you see both, you may set the two distances from any point on the bisector to the endpoints equal to each other.

The circumcenter

The three perpendicular bisectors of a triangle's sides are concurrent — they all pass through one point, the circumcenter. If $P$ is the circumcenter of $\triangle ABC$, then $PA = PB = PC$. That shared distance is the radius of the circle that passes through all three vertices, which is why the circumcenter is the center of the circumscribed circle.

The circumcenter's location depends on the type of triangle. In an acute triangle it sits inside the triangle. In a right triangle it lands exactly on the midpoint of the hypotenuse. In an obtuse triangle it falls outside the triangle. It does not have to be inside — that is a common trap answer.

The figure shows the circumcenter $P$ of an acute triangle: the three equal distances $PA$, $PB$, and $PC$ are the radius of the circle drawn through all three vertices.

Setting up the equation

Almost every algebra problem in this lesson uses the same move: identify two distances that the theorem forces to be equal, set the expressions equal, and solve for the variable. Then read the question again — it usually asks for a length, not for $x$, so substitute your $x$ back in before answering.

Worked examples

Try one yourself

Common questions