Divide a line segment of length 14 cm internally in the ratio 2 : 5. Also, justify your construction.

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 A₁, A₂, A₃, A₄, A₅, A₆, and A₇ such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇

Step 3: We join points A₇ and B.

Step 4: We draw line segment A₂P such that A₂P || A₇B and P is the point of intersection of this line segment with AB.

Point P divides AB in the ratio 2 : 5.

Justification –

In ΔAA₂P and ΔAA₇B,

v. ∠A is common.

vi. ∠AA₂P = ∠AA₇B (corresponding angles ∵ A₂P || A₇B)

Hence, ΔAA₂P ~ ΔAA₇B

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

⇒

Let AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇ = 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