Draw a triangle \(ABC\) of base \(BC = 8 \ cm\), \(\angle A = 60^{\circ}\) and the bisector of \(\angle A\) meets \(BC\) at \(D\) such that \(BD = 6 \ cm\).
Arrange the steps of construction in a proper order:
\(Step-1:\) Draw a line segment \(BC\) of length \(\ cm\).
\(Step-2: \) At \(B\), draw \(BE\) such that \(\angle CBE =\) \(^{\circ}\).
\(Step-3:\) At \(B\), draw \(BF\) such that \(\angle EBF =\) \(^{\circ}\).
\(Step-4:\) Draw the to \(BC\), which intersects \(BF\) at \(O\) and \(BC\) at \(G\).
\(Step-5:\) With \(O\) as centre and \(OB\) as radius draw a .
\(Step-6:\) From \(B\) mark an of \(6 \ cm\) on \(BC\) at \(D\).
\(Step-7:\) The intersects the circle at \(I\). Join \(ID\).
\(Step-8:\) \(ID\) produced meets the at \(A\). Now, join \(AB\) and \(AC\). Then, \(\triangle ABC\) is the required triangle.