Construct a triangle of sides $4 \, cm$,$5 \, cm$ and $6 \, cm$ and then a triangle similar to it whose sides are $\frac{2}{3}$ of the corresponding sides of the first triangle. Also,give the justification of the construction.

(N/A) $1.$ Draw a line segment $AB = 4 \, cm$. Taking point $A$ as centre,draw an arc of $5 \, cm$ radius. Similarly,taking point $B$ as its centre,draw an arc of $6 \, cm$ radius. These arcs will intersect each other at point $C$. Now,$AC = 5 \, cm$ and $BC = 6 \, cm$,and $\triangle ABC$ is the required triangle.
$2.$ Draw a ray $AX$ making an acute angle with line $AB$ on the opposite side of vertex $C$.
$3.$ Locate $3$ points $A_1, A_2, A_3$ on line $AX$ such that $AA_1 = A_1A_2 = A_2A_3$.
$4.$ Join $BA_3$ and draw a line through $A_2$ parallel to $BA_3$ to intersect $AB$ at point $B'$.
$5.$ Draw a line through $B'$ parallel to the line $BC$ to intersect $AC$ at $C'$.
$\triangle AB'C'$ is the required triangle.
The construction can be justified by proving that $\frac{AB'}{AB} = \frac{B'C'}{BC} = \frac{AC'}{AC} = \frac{2}{3}$.
By construction,we have $B'C' \parallel BC$.
$\therefore \angle AB'C' = \angle ABC$ (Corresponding angles).
In $\triangle AB'C'$ and $\triangle ABC$,
$\angle AB'C' = \angle ABC$ (Proved above),
$\angle B'AC' = \angle BAC$ (Common angle).
$\triangle AB'C' \sim \triangle ABC$ ($AA$ similarity criterion).
$\Rightarrow \frac{AB'}{AB} = \frac{B'C'}{BC} = \frac{AC'}{AC}$ .......... $(1)$
In $\triangle AA_2B'$ and $\triangle AA_3B$,
$\angle A_2AB' = \angle A_3AB$ (Common angle),
$\angle AA_2B' = \angle AA_3B$ (Corresponding angles).
$\therefore \triangle AA_2B' \sim \triangle AA_3B$ ($AA$ similarity criterion).
$\Rightarrow \frac{AB'}{AB} = \frac{AA_2}{AA_3} = \frac{2}{3}$ ......... $(2)$
From equations $(1)$ and $(2)$,we obtain $\frac{AB'}{AB} = \frac{B'C'}{BC} = \frac{AC'}{AC} = \frac{2}{3}$.
This justifies the construction.