I draw an obtuse angled triangle and let us prove that the perpendicular bisectors of sides are concurrent. Let us write where the circumcentre lies (inside/outside/on the side) of the triangular region.

Let us consider an obtuse angled triangle PQR.

OA, OB and OC perpendicularly bisect the sides.

Joined OP, OQ and OR!

In Δ AOQ and Δ AOR

AO = AO (Common side)

AQ = AR (A is the midpoint of QR)

∠OAQ = ∠OAR (OA is the perpendicular bisector)

Hence Δ QOA and Δ ROA are congruent to each other by S.A.S. axiom of congruency

Hence OQ = OR (Corresponding parts of Congruent Triangles)

In Δ QOC and Δ POC

CO = CO (Common side)

CQ = CP(C is the midpoint of QP)

∠OCQ = ∠OCP (OC is the perpendicular bisector)

Hence Δ QOC and Δ POC are congruent to each other by S.A.S. axiom of congruency

Hence OQ = OP (Corresponding parts of Congruent Triangles)

Hence OP = OQ = OR

So it is proved that the perpendicular bisectors are concurrent

The circum centre lies outside in an obtuse angled triangle