A triangle similar to given $\mathrm{Δ}ABC$with sides equal to $\frac{3}{4}$ of the sides of $\mathrm{Δ}ABC$ is to be constructed as the given image. Arrange the following steps of construction in 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'.