Steps of Construction:

Step I: BC = 5 cm is drawn.

Step II: Perpendicular bisector of BC is drawn and it intersects BC at O.

Step III: At a distance of 4 cm, a point A is marked on perpendicular bisector of BC.

Step IV: AB and AC is joined to form ΔABC.

Step V: A ray BX is drawn making an acute angle with BC opposite to vertex A.

Step VI: 3 points B1 B2 and B3 is marked BX.

Step VII: B2 is joined with C to form B2C.

Step VIII: B3C' is drawn parallel to B2C and C'A' is drawn parallel to CA.

Thus, A'BC' is the required triangle formed.

Justification:

ΔAB'C' ~ ΔABC by AA similarity condition.

∴ AB/AB' = BC/B'C' = AC/AC'

also,

AB/AB' = AA2/AA3 = 2/3

⇒ AB' = 3/2 AB, B'C' = 3/2 BC and AC' = 3/2AC