Steps of construction:

1. BC=4cm is drawn.

2. At B, a ray of 5 cm making an angle of 90° with BC is drawn at A. Join AC.

3. A ray BX is drawn making an acute angle with BC opposite to vertex A. Five points B₁ , B₂ , B₃, B₄, and B₅ at equal distance is marked on BX.

4. B₃C is joined B₅C₁ is made parallel to B₃C.

5. Now A₁C₁ is made parallel to AC.

Triangle A₁BC1 is the required triangle.

Justification:

Since the scale factor is ,

We need to prove,

By construction,

… (1)

Also, A₁C₁ is parallel to AC.

So, this will make same angle with BC.

∴ ∠A₁C₁B = ∠ACB …. (2)

Now,

In ΔA₁BC₁ and ΔABC

∠ B = ∠ B (common)

∠A₁C₁B = ∠ACB (from 2)

ΔA₁BC₁∼ ΔABC

Since corresponding sides of similar triangles are in same ratio.

From (1)

Hence construction is justified.