
Steps of Construction:
Step I: Draw BC = 6 cm.
Step II: Construct XY, the perpendicular bisector of line segment BC, meeting BC at M.
Step III: Along MP, cut-off MA = 4 cm.
Step IV: Join BA and CA. Then DABC so obtained is the required ∆ABC.
Step V: Extend BC to D, such that BD = 9 cm(=3/2x6cm).
Step VI: Draw DE || CA meeting BA produced at E. Then DEBD is the required triangle.
Justification:

Hence, we have the new triangle similar to the given triangle whose sides are 1 1/2 i.e,3/2 times the corresponding sides of the isosceles ∆ABC.