Given a line segment AB, we want to divide it in the ratio $m : n$, where both $m$ and $n$ are positive integers. To help you to understand it, we shall take $m$ = 3 and $n$ = 2.
1. Draw any ray AX, making an acute angle with AB.
2. Locate 5 (= $m + n$) points $A_1$, $A_2$,$A_3$,$A_4$ and $A_5$, on AX so that $A$$A_1$ = $A_1$$A_2$ = $A_2$$A_3$ = $A_3$$A_4$ = $A_4$$A_5$.
3. Join B$A_5$.
4. Through the point $A_3$ (m = 3), draw a line parallel to $A_5B$ (by making an angle equal to ∠AA B) at $A_3$, intersecting AB at the point C (see Fig. 11.1). Then, AC : CB = 3 : 2.
Let us see how this method gives us the required division.
Since $A_3C$ is parallel to $A_5B$, therefore,
${AA_3}/{A_3A_5}$=${AC}/{CB}$ ......... (By the Basic Proportionality Theorem)
By Consruction, ${AA_3}/{A_3A_5}$=$3/2$,
Therefore, ${AC}/{CB}$ =${3/2}$
This shows that C divides AB in the ratio $3 : 2$.
1. Draw any ray AX making an acute angle with AB.
2. Draw a ray BY parallel to AX by making ∠ ABY equal to ∠ BAX.
3. Locate the points $A_1$, $A_2$,$A_3$ ($m$ = 3) on AX and $B_1$, $B_2$ ($n$ = 2) on BY such that $AA_1$ = $A_1A_2$ = $A_2A_3$ = $BB_1$ = $B_1B_2$.
4. Join $A_3B_2$. Let it intersect $AB$ at a point $C$ (see Fig. 11.2). Then AC : CB = 3 : 2.
Then ${AA_3}/{BB_2}$=${AC}/{BC}$
Since , by construction, ${AA_3}/{BB_2}$=$3/2$
Therefore,${AC}/{BC}$=$3/2$
In fact, the methods given above work for dividing the line segment in any ratio. We now use the idea of the construction above for constructing a triangle similar to a given triangle whose sides are in a given ratio with the corresponding sides of the given triangle.