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 A1, A2, A3, A4 and A5 on AX so that AA1 = A1A2 = A2A3 = A3A4 = A4A5.
3. Join BA5.
4. Through the point A3 (m = 3), draw a line parallel to A5B (by making an angle equal to ∠AA5B) at A3 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 A3C is parallel to A5B, therefore,
This shows that C divides AB in the ratio 3 : 2.
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 A1, A2, A3 (m = 3) on AX and B1, B2 (n = 2) on BY such that AA1 = A1A2 = A2A3 = BB1 = B1B2.
4. Join A3B2.
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 DAA3C is similar to DAB2C. (Why ?)
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.