Give the justification of the construction also:
Construct a triangle with sides $5\, cm$,$6\, cm$,and $7\, cm$ and then another triangle whose sides are $\frac{7}{5}$ of the corresponding sides of the first triangle.

(N/A) $1.$ Draw a line segment $AB$ of $5\, cm$. Taking $A$ and $B$ as centers,draw arcs of $6\, cm$ and $7\, cm$ radius respectively. Let these arcs intersect each other at point $C$. $\triangle ABC$ is the required triangle having side lengths of $5\, cm$,$6\, cm$,and $7\, cm$ respectively.
$2.$ Draw a ray $AX$ making an acute angle with line $AB$ on the opposite side of vertex $C$.
$3.$ Locate $7$ points,$A_1, A_2, A_3, A_4, A_5, A_6, A_7$ (as $7$ is greater between $5$ and $7$) on line $AX$ such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5 = A_5A_6 = A_6A_7$.
$4.$ Join $BA_5$ and draw a line through $A_7$ parallel to $BA_5$ to intersect the extended line segment $AB$ at point $B'$.
$5.$ Draw a line through $B'$ parallel to $BC$ intersecting the extended line segment $AC$ at $C'$. $\triangle AB'C'$ is the required triangle.
Justification:
The construction can be justified by proving that $AB' = \frac{7}{5} AB$,$B'C' = \frac{7}{5} BC$,and $AC' = \frac{7}{5} AC$.
In $\triangle ABC$ and $\triangle AB'C'$:
$\angle ABC = \angle AB'C'$ (Corresponding angles)
$\angle BAC = \angle B'AC'$ (Common)
$\triangle ABC \sim \triangle AB'C'$ ($AA$ similarity criterion)
$\Rightarrow \frac{AB}{AB'} = \frac{BC}{B'C'} = \frac{AC}{AC'} = \frac{5}{7}$ (Equation $1$)
In $\triangle AA_5B$ and $\triangle AA_7B'$:
$\angle A_5AB = \angle A_7AB'$ (Common)
$\angle AA_5B = \angle AA_7B'$ (Corresponding angles)
$\triangle AA_5B \sim \triangle AA_7B'$ ($AA$ similarity criterion)
$\Rightarrow \frac{AB}{AB'} = \frac{AA_5}{AA_7} = \frac{5}{7}$ (Equation $2$)
On comparing equations $1$ and $2$,we obtain:
$\frac{AB}{AB'} = \frac{BC}{B'C'} = \frac{AC}{AC'} = \frac{5}{7}$
$\Rightarrow AB' = \frac{7}{5} AB, B'C' = \frac{7}{5} BC, AC' = \frac{7}{5} AC$.
This justifies the construction.