Steps for construction:

i) Draw a line segment AB = 6 cm

ii) Draw a ray at A, making an angle of 60^{o}_{ }with BC.

iii) With B as centre and radius = 6.2 cm draw an arc which intersects AX ray at C.

iv) Join BC.

Then, ΔABC is the required triangle.

v) Draw the perpendicular bisectors of AB and AC intersecting each other at O.

vi) With centre O, and radius as OA or OB or OC, draw a circle which will pass through A, B and C.

vii) From O, draw OD ⊥ AB.

Proof: In right ΔOAD and ΔOBD

OA = OB [Radii of same circle]

OD = OD [Common]

∆OAD ≅ ∆OBD by RHS congruence criterion.

Hence, by CPCT

AD = BD