Question

Divide a line segment of length $10cm$ internally in the ratio $3:2.$

Answer 1

We follow the following step of construction.

Steps of construction
$StepI$
Drawn a line segment $AB=10cm$ by using a ruler.

$StepII$
Drawn any ray making an acute angle $\angle BAX$ with $AB$.

$StepIII$
Along $AX$, mark-off $5(=3+2)$ points $A_1,A_2,A_3,A_4$ and $A_5$ such that
${AA}_1=A_1{A}_2=A_3{A}_4=A_4{A}_5$.

$StepIv$
Join ${BA}_5$

$Stepv$
Through $A_3$ draw a line $A_3$ $P$ parallel to $A_5$ $B$ by making an angle equal to $\angle {AA}_{5}B$ at $A_3$ intersecting $AB$ at a point $P$.

The point $P$ so obtained is the required point, which divides $AB$ internally in the ration $3:2$.

ALTERNATIVE METHOD FOR DIVISION OF A LINE SEGMENT INTERNALLY IN A GIVEN RATIO

We may use the following steps to divide a given line segment $AB$ internally in a given ratio $m:n$, where $m$ and $n$ are natural numbers.

Steps of construction

$StepI$
Draw line segment $AB$ of given length.

$StepII$
Draw any ray $AX$ making an acute angle $\angle BAX$ with $AB$.

$StepIII$
Draw a ray $BY$, on opposite side of $AX$, parallel to $AX$ by making an angle $\angle BAY$ equal to $\angle BAX$.

$StepIV$
Mark off $m$ points $A_1,A_2,,A_m$, on $AX$ and $n$ points $B_1,B_2,,B_n$ on $BY$ such that
${AA}_1=A_1{A}_2=.=A_{m-1}{A}_m$
$={BB}_1=B_1{B}_2=.=Bn-1B_n$.

$StepV$
Join $A_m{B}_n$. Suppose it intersects $AB$ at $P$.

The point $P$ is the required point dividing $AB$ in the ratio $m:n$.

In triangles ${AA}_{m}P$ and ${BB}_{n}P$, we have

$\angle A_{m}AP=\angle PB{B}_n$ and, $AP{A}_m=\angle BP{B}_n$

So, by $AA$ similarly criterion, we have

$△A{A}_{m}P-△B{B}_{n}P$

$\Rightarrow \frac{{AA}_m}{{BB}_n}=\frac{AP}{BP}$

$\Rightarrow \frac{AP}{BP}=\frac{m}{n}$