Step 1
Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5cm radius Similarly, taking point B as its centre, draw an arc of 6 cm and ΔABC is the required triangle.
Step 2
Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.
Step 3
Locate 3 points A₁, A₂, A₃ (as 3 is greater between 2 and 3) on line AX such that AA₁ = A₁A₂ = A₂A₃.
Step 4
Join BA₃ and draw a line through A₂ parallel to BA₃ to intersect AB at point B’
Step 5
Draw a line through B’ parallel to the line BC to intersect AC at C’.
ΔAB’C is the required triangle.
Justification
The construction van be justified by proving that
AB’ = 2/3 AB, B’C’ = 2/3 BC, AC’ = 2/3 AC
By construction, we have B’C’ II BC
∴ ∠AB’C’ = ∠ABC (corresponding angles)
In ΔAB’C’ and ΔABC,
∠AB’C’ = ∠ABC (Proved above)
∠B’AC’ = ∠ BAC (common)
∴ Δ AB’C’ ∼ ΔABC (AA similarity criterion)
⇒ AB’/AB = B’C’/BC = AC/AC …(1)
In Δ AA₂B’ and ΔAA₃B,
∠A₂AB’ = ∠A₃AB (common)
∠AA₂B’ = ∠AA₃B (corresponding angles)
∴ ΔAA₂B’ ∼ ΔAA₃B (AA similarity criterion)
⇒ AB’/AB = AA₂/AA₃
⇒ AB’/AB = 2/3 …(2)
From equations (1) and (2), we obtain
AB’/AB = B’C’/BC = AC’/AC = 2/3
⇒ AB’ = 2/3 AB, B’C’ = 2/3 BC, AC’ = 2/3 AC
This justifies the construction.