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 △ABC so obtained is the required △ABC.
Step V: Extend BC to D, such that BD = 9 cm (=3/2 x 6 cm).
Step VI: Draw DE || CA meeting BA produced at E. Then △EBD is the required triangle.
Justification:
Since DE || CA
△ABC ~ △EBD and EB/AB = DE/CA = BD/BC =9/6 =3/2
Hence, we have the new triangle similar to the given triangle whose sides are 3/2 times the corresponding sides of the isosceles △ABC.
