Answer :

We need to divide this line segment AB of length 12 cm internally in the ratio 2 : 3.

__Step 1:__ Draw a line segment AC of arbitrary length and at an any angle to AB such that ∠CAB is acute.

__Step 2:__ We plot (2 + 3 =) 5 points A_{1}, A_{2}, A_{3}, A_{4}, and A_{5} such that AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5}.

__Step 3:__ We join points A_{5} and B.

__Step 4:__ We draw line segment A_{2}P such that A_{2}P || A_{5}B and P is the point of intersection of this line segment with AB.

Point P divides AB in the ratio 2 : 3.

__Justification__ –

In ΔAA_{2}P and ΔAA_{5}B,

i. ∠A is common.

ii. ∠AA_{2}P = ∠AA_{5}B (corresponding angles ∵ A_{2}P || A_{5}B)

Hence, ΔAA_{2}P ~ ΔAA_{5}B

So, ratio of lengths of corresponding sides must be equal.

⇒

Let AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5} = x

So, the previous relation can be re – written as –

⇒ 2(AP +PB) = 5AP

⇒ 2PB = 3AP

⇒ AP/PB = 2/3, or, AP : PB = 2 : 3

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