A triangle ABC with sides BC =7cm, $\angle B={45}^{\circ },\angle A={105}^{\circ }$ is given. The image of constructing a similar triangle of $\Delta ABC$ whose sides are $\frac{3}{4}$ times the corresponding sides of the $\Delta ABC$ is given below. Arrange the steps of construction in the correct order.
1.Draw any ray BX making an acute angle with BC on the side opposite to  the vertex A.
2.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'.
3.Draw a line through C' parallel to the line CA to intersect BA at A'.
4.Locate 4(the greater of 3 and 4 in$\frac{3}{4}$) points $B_1,B_2,B_3\text{and}{B}_4$ on BX  so that $B{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4$