Determine a point which divides a line segment of length 12 cm internally in the ratio 2 : 3. Also, justify your construction.

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 A₁, A₂, A₃, A₄, and A₅ such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.

Step 3: We join points A₅ and B.

Step 4: We draw line segment A₂P such that A₂P || A₅B and P is the point of intersection of this line segment with AB.

Point P divides AB in the ratio 2 : 3.

Justification –

In ΔAA₂P and ΔAA₅B,

i. ∠A is common.

ii. ∠AA₂P = ∠AA₅B (corresponding angles ∵ A₂P || A₅B)

Hence, ΔAA₂P ~ ΔAA₅B

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

⇒

Let AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = 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