Answer :

We need to divide this line segment AB of length 14 cm internally in the ratio 2 : 5.

__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 + 5 =) 7 points A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, and A_{7} such that AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5} = A_{5}A_{6} = A_{6}A_{7}

__Step 3:__ We join points A_{7} and B.

__Step 4:__ We draw line segment A_{2}P such that A_{2}P || A_{7}B and P is the point of intersection of this line segment with AB.

Point P divides AB in the ratio 2 : 5.

__Justification__ –

In ΔAA_{2}P and ΔAA_{7}B,

v. ∠A is common.

vi. ∠AA_{2}P = ∠AA_{7}B (corresponding angles ∵ A_{2}P || A_{7}B)

Hence, ΔAA_{2}P ~ ΔAA_{7}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} = A_{5}A_{6} = A_{6}A_{7} = x

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

⇒ 2(AP +PB) = 7AP

⇒ 2PB = 5AP

⇒ AP/PB = 2/5, or, AP : PB = 2 : 5

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