Draw a line segment of length $7 \, cm$. Find a point $P$ on it which divides it in the ratio $3: 5$.

(N/A) $1.$ Draw a line segment $AB = 7 \, cm$.
$2.$ Draw a ray $AX$ making an acute angle $\angle BAX$.
$3.$ Along $AX,$ mark $3 + 5 = 8$ points $A_1, A_2, A_3, A_4, A_5, A_6, A_7, A_8$ such that $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5 = A_5A_6 = A_6A_7 = A_7A_8$.
$4.$ Join $A_8B$.
$5.$ From $A_3,$ draw $A_3P \parallel A_8B$ meeting $AB$ at $P$ (by making an angle equal to $\angle BA_8A$ at $A_3$).
Then,$P$ is the point on $AB$ which divides it in the ratio $3: 5.$
Thus,$AP: PB = 3: 5.$
Justification:
Let $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = \dots = A_7A_8 = x.$
In $\triangle ABA_8,$ we have $A_3P \parallel A_8B.$
By Basic Proportionality Theorem,$\frac{AP}{PB} = \frac{AA_3}{A_3A_8} = \frac{3x}{5x} = \frac{3}{5}.$
Hence,$AP: PB = 3: 5.$