Divide a line segment of length 9 cm internally in the ratio 4 : 3. Also, give justification of the construction.

We need to divide this line segment AB of length 9 cm internally in the ratio 4 : 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 (4 + 3 =) 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 A₄P such that A₄P || A₇B and P is the point of intersection of this line segment with AB.

Point P divides AB in the ratio 4 : 3.

Justification –

In ΔAA₄P and ΔAA₇B,

iii. ∠A is common.

iv. ∠AA₄P = ∠AA₇B (corresponding angles ∵ A₄P || A₇B)

Hence, ΔAA₄P ~ 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 –

⇒ 4(AP +PB) = 7AP

⇒ 4PB = 3AP

⇒ AP/PB = 4/3, or, AP : PB = 4 : 3