To draw a parallelogram ABCD in which BC = 5 cm, AB = 3 cm, and angle ABC = 60 degree, we need the following steps:

1. Draw a line segment AB = 3 cm

2. At B, construct angle ABM = 60 degree

3. From Bm, cut a line segment BC = 5 cm

4. Now, from point A, draw AN || BC

5. From the point C, draw CD || BA, meeting AN at D

So, ABCD is the required parallelogram.

Now, from diagonal BD, divide the parallelogram ABCD into two triangles BCD and ABD

To construct the triangle BD₁ C₁ similar to triangle BDC with scale factor 4 : 3, we do the following steps:

1. Below BD, make an acute angle DBX

2. Along BX, make four points B₁ , B₂ ,B₃ , B₄ such that BB₁ , B₁ B₂ , B₂ B₃ , B₃ B₄

3. Join B₃ D

4. From B₄ , draw B₃ D₁ || B₃ D meeting BD produced at D₁

5. From D₁ , draw C₁ D₁ || CD meeting BD produced at C₁

Thus B D₁ C₁ is the required triangle whose sides are 4/3 times the corresponding side of triangle BDC.