Draw the bisector of `/_AOB=60^(@)`.

Explanation

(a) Draw `/_AOB=60^(@)`.
(b) With O as the centre and some convenient radius draw an arc which intersects `bar(OA) ` and `bar(OB)` at X and Y respectively.
(c ) With X as the centre and more than half of XY as radius an arc.
(d) With Y as the centre and with the same radius draw an arc which cuts the previous arc at P.
(e ) Join `bar(OP)`. `bar(OP)` is the bisector of `/_AOB`.
`i.e., /_POB=/_ POA =30^(@)`