A ΔA’BC’ whose sides are 3/4 of the corresponding sides of Δ ABC can be drawn as follows.
Step 1
Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
Step 2
Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
Step 3
Locat 4 points (as 4 is greater in 3 and 4), B₁, B₂, B₃, B₄, on line segment BX.
Step 4
Join B₄C and Draw a line through B₃, parallel to B₄C intersecting BC AT C’.
Step 5
Draw a line through C’ parallel to AC intersecting AB at A’. ΔA’BC’ is the required triangle.
Justification
The construction can be justified by proving A’B = 3/4 AB, BC’ = 3/4 BC, A’C’ = 3/4 AC
In ΔA’BC’ and ΔABC,
∠A’C’B = ∠ACB (Corresponding angles)
∠A’BC’ = ∠ABC (Common)
∴ ΔA’BC’ ∼ ΔABC (AA similarity criterion)
⇒ A’B/AB = BC’/BC = A’C’/AC …(1)
In ΔBB₃C’ and ΔBB₄C,
∠B₃BC’ = ∠B₄BC (Common)
∠BB₃C’ = ∠BB₄C ( Corresponding angles)
∴ ΔBB₃C ∼ ΔBB₄C (AA similarity criterion)
⇒ BC’/BC = BB₃/BB₄ ⇒ BC’/BC = 3/4 …(2)
From equations (1) and (2), we obtain
A’B/AB = BC’/BC = A’C’/AC = 3/4
⇒ A’B = 3/4 AB, BC’ = 3/4 BC A’C’ = 3/4 AC
This justifies the construction.