### Theory:

Let us see the alternative method to divide a line segment in a given ratio.
Example:
Draw a line segment of length $$6.2 \ cm$$ and divide it in the ratio $$4:3$$. Measure the two parts.

Step 1: Draw a line segment $$AB$$ of length $$6.2 \ cm$$.

Step 2: Draw any ray $$AX$$, making an acute angle with $$AB$$.

Step 3: Mark $$4$$ points $$A_1$$, $$A_2$$, $$A_3$$ and $$A_4$$ on $$AX$$ such that $$AA_1$$ $$= A_1A_2$$ $$= A_2A_3$$ $$= A_3A_4$$.

Step 4: Draw another ray $$BY$$ parallel to $$AX$$ and mark $$3$$ points on $$BY$$ such that $$BB_1$$ $$= B_1B_2$$ $$= B_2B_3$$.

Step 5: Join $$A_4$$ to $$B_3$$. Let it intersect at $$C$$ on $$AB$$. Thus, $$AC: CB = 4:3$$

JUSTIFICATION:

In $$\Delta AA_4C$$ and $$\Delta BB_3C$$:

$$CAA_4 = CBB_3$$ (Since $$AX || BY$$, the alternate interior angles are equal)

$$ACA_4 = BCC_3$$ (Vertically opposite angles are equal)

Therefore, $$\Delta AA_4C \sim \Delta BB_3C$$ (by AA similarity)

If two triangles are similar then their corresponding sides are in the same ratio.

$\frac{A{A}_{4}}{B{B}_{3}}=\frac{\mathit{AC}}{\mathit{BC}}$

By construction, we have:

$\frac{A{A}_{4}}{B{B}_{3}}=\frac{4}{3}$

So, $\frac{\mathit{AC}}{\mathit{BC}}=\frac{4}{3}$.

Therefore, $$AC : BC = 4: 3$$.