Draw a right triangle $ABC$ in which $BC = 12 \, cm$,$AB = 5 \, cm$ and $\angle B = 90^{\circ}$. Construct a triangle similar to it with a scale factor of $\frac{2}{3}$. Is the new triangle also a right triangle?

(A) Steps of construction:
$1.$ Draw a line segment $BC = 12 \, cm$.
$2.$ From point $B$,draw a line $AB = 5 \, cm$ perpendicular to $BC$ such that $\angle B = 90^{\circ}$.
$3.$ Join $AC$. Thus,$\triangle ABC$ is the given right triangle.
$4.$ From $B$,draw an acute angle $\angle CBY$ downwards.
$5.$ On ray $BY$,mark three points $B_1, B_2$ and $B_3$ such that $BB_1 = B_1B_2 = B_2B_3$.
$6.$ Join $B_3C$.
$7.$ From point $B_2$,draw $B_2N \parallel B_3C$ to intersect $BC$ at $N$.
$8.$ From point $N$,draw $NM \parallel CA$ to intersect $BA$ at $M$. Thus,$\triangle MBN$ is the required triangle.
$9.$ Since $NM \parallel CA$ and $BC$ is a transversal,the corresponding angles are equal. Therefore,$\angle MNB = \angle ACB$. Since $\angle B = 90^{\circ}$ in both triangles,$\triangle MBN$ is also a right-angled triangle at $B$.