In any triangle the perpendicular bisectors of the sides are concurrent.

In $\mathrm{△}ABC,D,E,F$ are mid points of the sides $BC,CA,AB$ respectively. Draw perpendiculars through $D,E$ meet at $O.$

$\therefore OD\perp BC,OE\perp AC$

Now we have to show $OF\perp AB$

Let $OA=OB=OC=R$ 

Now, $\overline{OF}\cdot \overline{AB}=\left(\frac{\overline{OB}+\overline{OA}}{2}\right)\cdot (\overline{OB}-\overline{OA})$

$=\frac{1}{2}\left({\overline{OB}}^2-{\overline{OA}}^2\right)=\frac{1}{2}\left(O{B}^2-O{A}^2\right)$

$=\frac{1}{2}\left(R^2-R^2\right)=0\therefore OF\perp AB$

Thus the perpendicular bisectors of a $\mathrm{△}ABC$ are concurrent.