**Division Of A Line Segment Into A Given Ratio**

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.

**Steps of Construction:**

**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 AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5}.

**3. ** Join BA_{5}.

**4.** Through the point A_{3} (m = 3), draw a line parallel to A_{5}B (by making an angle equal to ∠AA_{5}B) at A_{3} intersecting AB at the point C (see figure). Then, AC : CB = 3 : 2.

Let use see how this method gives us the required division.

Since A_{3}C is parallel to A_{5}B, therefore,

\( \frac{A{{A}_{3}}}{{{A}_{3}}{{A}_{5}}}=\frac{AC}{CB}\text{ }\left( \text{By the Basic Proportionality Theorem} \right) \)

\( \frac{A{{A}_{3}}}{{{A}_{3}}{{A}_{5}}}=\frac{3}{2}\text{ (By construction) } \)

\( \text{ }\frac{AC}{CB}=\frac{3}{2}\text{ } \)

This shows that C divides AB in the ratio 3 : 2.

**Alternative Method **

**Steps of Construction : **

**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_{1}A_{2} = A_{2}A_{3} = BB_{1} = B_{1}B_{2}.

**4.** Join A_{3}B_{2}.

Let it in intersect AB at a point C (see figure)

Then AC : CB = 3 : 2

Whey does this method work ? Let us see.

Here DAA_{3}C is similar to DAB_{2}C. (Why ?)

\( \text{Then }\frac{A{{A}_{3}}}{B{{B}_{2}}}=\frac{AC}{BC}\)

\( \frac{A{{A}_{3}}}{B{{B}_{2}}}=\frac{3}{2}\text{ (By construction) } \)

\( \text{ }\frac{AC}{BC}=\frac{3}{2} \)

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.