Perpendicular Bisectors of Triangle Meet at Point

Theorem
Let $\triangle ABC$ be a triangle.

The perpendicular bisectors of $AB$, $BC$ and $AC$ all intersect at the same point.

Proof
Let the perpendicular bisectors of $AC$ and $AB$ be constructed through $D$ and $E$ respectively to meet at $F$.


 * PerpendicularBisectorsMeetAtPoint.png

By definition of perpendicular bisector:
 * $AE = EB$

and:
 * $\angle AEF = \angle BEF$ are right angles.

From Triangle Side-Angle-Side Equality:
 * $\triangle AEF = \triangle BEF$

and so $AF = BF$.

Similarly, by definition of perpendicular bisector:
 * $AD = DCB$

and:
 * $\angle ADF = \angle CDF$ are right angles.

From Triangle Side-Angle-Side Equality:
 * $\triangle ADF = \triangle CDF$

and so $AF = CF$.

Thus:
 * $BF = CF$

Let $FG$ be the angle bisector of $\angle BFC$.

We have:
 * $BF = CF$ from above

and:
 * $\angle BFG = \angle CFG$ by construction

Thus by Triangle Side-Angle-Side Equality:
 * $\triangle BFG = \triangle CFG$

and so $BG = CG$.

Thus as $\angle BGF = \angle CGF$ it follows that they are both right angles.

Thus $FG$ is the perpendicular bisector of $BC$.

Thus we have all three perpendicular bisectors of the sides of $ABC$ meeting at the same point $G$.

Also see

 * Circumscribing Circle about Triangle, where it can be seen that $G$ is the circumcenter of $\triangle ABC$.