Draw a triangle $ABC$ in which $AB = 4 \, cm$,$BC = 6 \, cm$,and $AC = 9 \, cm$. Construct a triangle similar to $\triangle ABC$ with a scale factor of $\frac{3}{2}$. Justify the construction. Are the two triangles congruent? Note that all three angles of the two triangles are equal,but the sides are not.

(N/A) Steps of construction:
$1.$ Draw a line segment $BC = 6 \, cm$.
$2.$ Taking $B$ and $C$ as centers,draw two arcs of radii $4 \, cm$ and $9 \, cm$ respectively,intersecting each other at $A$.
$3.$ Join $BA$ and $CA$. $\triangle ABC$ is the required triangle.
$4.$ From $B$,draw any ray $BX$ downwards making an acute angle.
$5.$ Mark three points $B_1, B_2, B_3$ on $BX$ such that $BB_1 = B_1B_2 = B_2B_3$.
$6.$ Join $B_2C$ and from $B_3$,draw $B_3M \parallel B_2C$ intersecting the extended line segment $BC$ at $M$.
$7.$ From point $M$,draw $MN \parallel CA$ intersecting the extended line segment $BA$ at $N$.
Then,$\triangle NBM$ is the required triangle whose sides are $\frac{3}{2}$ times the corresponding sides of $\triangle ABC$.
The two triangles are not congruent because congruent triangles must have the same shape and size. Here,the triangles are similar (same shape),but their sizes are different.
Justification:
Since $B_3M \parallel B_2C$,by Basic Proportionality Theorem:
$\frac{BC}{CM} = \frac{BB_2}{B_2B_3} = \frac{2}{1}$.
Now,$\frac{BM}{BC} = \frac{BC + CM}{BC} = 1 + \frac{CM}{BC} = 1 + \frac{1}{2} = \frac{3}{2}$.
Also,$MN \parallel CA$,therefore $\triangle ABC \sim \triangle NBM$.
Thus,$\frac{NB}{AB} = \frac{NM}{AC} = \frac{BM}{BC} = \frac{3}{2}$.