Construct a triangle similar to a given triangle $ABC$ with its sides equal to $\frac{5}{3}$ of the corresponding sides of the triangle $ABC$ (i.e.,of scale factor $\frac{5}{3}$).

(N/A) Given a triangle $ABC$,we are required to construct a triangle whose sides are $\frac{5}{3}$ of the corresponding sides of $\Delta ABC$.
Steps of Construction:
$1.$ Draw any ray $BX$ making an acute angle with $BC$ on the side opposite to the vertex $A$.
$2.$ Locate $5$ points (the greater of $5$ and $3$ in $\frac{5}{3}$) $B_{1}, B_{2}, B_{3}, B_{4}$ and $B_{5}$ on $BX$ so that $BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5}$.
$3.$ Join $B_{3}$ (the $3$rd point,$3$ being smaller of $3$ and $5$ in $\frac{5}{3}$) to $C$ and draw a line through $B_{5}$ parallel to $B_{3}C$,intersecting the extended line segment $BC$ at $C^{\prime}$.
$4.$ Draw a line through $C^{\prime}$ parallel to $CA$ intersecting the extended line segment $BA$ at $A^{\prime}$.
Then $\Delta A^{\prime}BC^{\prime}$ is the required triangle.
For justification of the construction,note that $\Delta ABC \sim \Delta A^{\prime}BC^{\prime}$.
Therefore,$\frac{AB}{A^{\prime}B} = \frac{AC}{A^{\prime}C^{\prime}} = \frac{BC}{BC^{\prime}}$.
But,$\frac{BC}{BC^{\prime}} = \frac{BB_{3}}{BB_{5}} = \frac{3}{5}$.
So,$\frac{BC^{\prime}}{BC} = \frac{5}{3}$,and therefore,$\frac{A^{\prime}B}{AB} = \frac{A^{\prime}C^{\prime}}{AC} = \frac{BC^{\prime}}{BC} = \frac{5}{3}$.