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 $F$.
By definition of perpendicular bisector:
- $AE = EB$
and:
- $\angle AEF = \angle BEF$ are right angles.
From Triangle Side-Angle-Side Congruence:
- $\triangle AEF = \triangle BEF$
and so $AF = BF$.
Similarly, by definition of perpendicular bisector:
- $AD = DC$
and:
- $\angle ADF = \angle CDF$ are right angles.
From Triangle Side-Angle-Side Congruence:
- $\triangle ADF = \triangle CDF$
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 Congruence:
- $\triangle BFG = \triangle CFG$
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 $G$.
$\blacksquare$
Also see
- Circumscribing Circle about Triangle, where it can be seen that $F$ is the circumcenter of $\triangle ABC$.
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {IV}$. Pure Geometry: Plane Geometry: The circumcentre
- 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): bisector or bisectrix