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,
( frac{A}}=frac{AC}{CB}text{ }left( text{By the Basic Proportionality Theorem} right) )
( frac{A}}=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.
[youtube https://www.youtube.com/watch?v=m4PSLIMfTiU?feature=oembed]
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 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 ?)
( text{Then }frac{A}{B}=frac{AC}{BC})
( frac{A}{B}=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.
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