Construct a triangle similar to a given triangle $ABC$ with its sides equal to $\frac{3}{4}$ of the corresponding sides of the triangle $ABC$ (i.e.,of scale factor $\frac{3}{4}$).

(N/A) Given a triangle $ABC$,we are required to construct another triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the triangle $ABC$.
Steps of Construction:
$1.$ Draw any ray $BX$ making an acute angle with $BC$ on the side opposite to the vertex $A$.
$2.$ Locate $4$ (the greater of $3$ and $4$ in $\frac{3}{4}$) points $B_1, B_2, B_3$ and $B_4$ on $BX$ so that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$.
$3.$ Join $B_4C$ and draw a line through $B_3$ (the $3rd$ point,$3$ being smaller of $3$ and $4$ in $\frac{3}{4}$) parallel to $B_4C$ to intersect $BC$ at $C'$.
$4.$ Draw a line through $C'$ parallel to the line $CA$ to intersect $BA$ at $A'$.
Then,$\Delta A'BC'$ is the required triangle.
Let us now see how this construction gives the required triangle.
Since $B_3C' \parallel B_4C$,by Basic Proportionality Theorem,$\frac{BC'}{C'C} = \frac{BB_3}{B_3B_4} = \frac{3}{1}$.
Therefore,$\frac{BC}{BC'} = \frac{BC' + C'C}{BC'} = 1 + \frac{C'C}{BC'} = 1 + \frac{1}{3} = \frac{4}{3}$,i.e.,$\frac{BC'}{BC} = \frac{3}{4}$.
Also,$C'A' \parallel CA$. Therefore,$\Delta A'BC' \sim \Delta ABC$.
So,$\frac{A'B}{AB} = \frac{A'C'}{AC} = \frac{BC'}{BC} = \frac{3}{4}$.