A triangle $ABC$ with $BC=6\text{Â}cm,AB=5\text{Â}cm$ and $\mathrm{âˆ}ABC={60}^{∘}$. The image of constructing a similar triangle of $\mathrm{Δ}ABC$ whose sides are $\frac{3}{4}$ times the corresponding sides of the $\mathrm{Δ}ABC$ is given below.

Arrange the steps of construction in the correct order.
1. Join $B_{4}C$ and raw a line through $B_3$
(the third point, 3 being smaller of 4 in $\frac{3}{4}$) parallel to $B_{4}C$  to intersect BC at C'.
2. Draw any ray $BX$ making an acute angle with BC on the side opposite to  the vertex A.
3. Locate 4 (the greater of 3 and 4 in$\frac{3}{4}$) points $B_1,B_2,B_3\text{Â}\text{and}\text{Â}{B}_4$ on BX  so that $B{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4$
4. Draw a line through C' parallel to the line CA to intersect BA at A'.