Why Perpendicular Bisectors Meet at One Point

An animated proof showing that all three perpendicular bisectors of a triangle intersect at a single point — the circumcenter — which is equidistant from all vertices.

Keywords: circumcenter, perpendicular bisector, equidistant, circumcircle, triangle centers

Prerequisites: perp-bisect-proof · Difficulty: beginner

The three perpendicular bisectors of a triangle's sides all pass through a single point, the circumcenter. A perpendicular bisector passes through a side's midpoint at a $90°$ angle.

Strategy: draw two of the bisectors and call their intersection $O$. Since $O$ lies on both, it is equidistant from all three vertices, which forces it onto the third bisector and lets us draw the circumcircle.

Step 1: Draw Two Perpendicular Bisectors

We start with triangle $ABC$ and draw the perpendicular bisectors of two sides.

The perpendicular bisector of a segment passes through its midpoint at a $90°$ angle. These two bisectors must intersect somewhere — let's call that point $O$.

Step 2: $O$ is Equidistant from All Vertices

Since $O$ lies on the perpendicular bisector of $AB$:

• $O$ is equidistant from $A$ and $B$
• So $OA=OB$

Since $O$ also lies on the perpendicular bisector of $BC$:

• $O$ is equidistant from $B$ and $C$
• So $OB=OC$

Combining these: $OA=OB=OC$. Point $O$ is equidistant from all three vertices!

Step 3: $O$ Lies on the Third Bisector

We've shown that $OA=OC$. This means $O$ is equidistant from $A$ and $C$.

But any point equidistant from two points must lie on their perpendicular bisector!

Therefore, $O$ must lie on the perpendicular bisector of $CA$ — the third perpendicular bisector passes through $O$ as well!

Step 4: The Circumcircle

Since $OA=OB=OC$, we can draw a circle centered at $O$ that passes through all three vertices.

This is the circumcircle — the unique circle passing through all three vertices of the triangle. And $O$ is the circumcenter.

All three perpendicular bisectors meet at the circumcenter $O$, which is equidistant from all three vertices:

Since $O$ is the same distance from all vertices, we can draw a circle centered at $O$ that passes through $A$, $B$, and $C$. This is the circumcircle, and $O$ is its center. ∎

Notes

Any point on the perpendicular bisector of a segment is equidistant from the segment's endpoints. This is the key fact that drives the entire proof.

Why? If you fold the paper along the perpendicular bisector, one endpoint lands exactly on the other, so any point on the fold line is the same distance from both.