The correct option is
A CircumcenterStep (I) A line BC=6cm is drawn.
Step (II) ∠ABC=60o&∠ACB=45o are drawn.
Step (III) OD, OE & OF are drawn as right bisectors of AB, BC & CA respectively.
The right bisector intersect at O.
So O is the circumcentre of ΔABC.
A circle is drawn with O as centre and either of OA, OB or OC as radius,
This is the circumcircle ofΔABC.
Justification-
Between the triangles OAF & OCF,
AF=FC (F is the mid point of AC),
OF is the common side,
∠OFA=∠OFC(OF&isrightbisectorofAC)∴ΔOAF≡ΔOCF.∴OA=OC..........(i)
Again, between ΔAOD&ΔBOD
AD=BD(DisthemidpointofAB),ODisthecommonside.∠ODA=∠ODB=90o∴ΔAOD≡ΔBOD.∴OA=OB........(ii).
Comparing (i) & (ii),
OA=OB=OC.
So we can draw a circle with o as centre and OA or OB or OC as radius.
So O is the circumcentre of ΔABC.