Draw a triangle $ABC$ in which $AB = 5 \, cm$,$BC = 6 \, cm$ and $\angle ABC = 60^{\circ}$. Construct a triangle similar to $\triangle ABC$ with a scale factor of $\frac{5}{7}$. Justify the construction.

(N/A) Steps of construction:
$1.$ Draw a line segment $AB = 5 \, cm$.
$2.$ At point $B$,construct $\angle ABY = 60^{\circ}$ and cut off $BC = 6 \, cm$ on $BY$.
$3.$ Join $AC$. Thus,$\triangle ABC$ is the required triangle.
$4.$ Draw any ray $AX$ making an acute angle with $AB$ on the side opposite to vertex $C$.
$5.$ Locate $7$ points $B_1, B_2, B_3, B_4, B_5, B_6, B_7$ on $AX$ such that $AB_1 = B_1B_2 = B_2B_3 = B_3B_4 = B_4B_5 = B_5B_6 = B_6B_7$.
$6.$ Join $B_7B$. Through $B_5$,draw a line parallel to $B_7B$ intersecting $AB$ at $M$.
$7.$ Through $M$,draw a line parallel to $BC$ intersecting $AC$ at $N$.
Then,$\triangle AMN$ is the required triangle whose sides are $\frac{5}{7}$ of the corresponding sides of $\triangle ABC$.
Justification:
Since $MN \parallel BC$,by Basic Proportionality Theorem,$\triangle AMN \sim \triangle ABC$.
By construction,$B_5M \parallel B_7B$. In $\triangle ABB_7$,by Thales theorem:
$\frac{AM}{AB} = \frac{AB_5}{AB_7} = \frac{5}{7}$.
Since $\triangle AMN \sim \triangle ABC$,the ratio of their corresponding sides is $\frac{AM}{AB} = \frac{AN}{AC} = \frac{MN}{BC} = \frac{5}{7}$.