Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ ABC = 60°. Then construct a triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the triangle ABC.

Steps of constructions:

1. Draw a line BC = 6 cm.
2. At B, make ∠C = 60° and cut an arc at A on the same line so that BA = 5 cm. Join AC, ΔABC is obtained.
3. Draw the ray BX such that ∠CBX is acute.
4. Mark 4 (since 4 > in 3/4) points B₁, B₂, B₃, B₄ on BX such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄
5. Join B₄ to C and draw B₃C' parallel to B₄C to intersect BC at C'.
6. Draw C'A' || CA to intersect BA at A’.

Now, ΔA'BC' is the required triangle similar to ΔABC where BA'/BA = BC'/BC = C'A'/CA = 3/4

Proof:

In ΔBB₄C , B₃C' || B₄C

Hence by Basic proportionality theorem,

B₃B₄/BB₃ = C'C/BC' = 1/3

C'C /BC' + 1 = 1/3 + 1

(C'C + BC')/BC' = 4/3

BC/BC' = 4/3 or BC'/BC = 3/4

Consider ΔBA'C' and ΔBAC

∠A'BC' = ∠ABC = 60°

∠BCA' = ∠BCA (Corresponding angles ∵ C'A'||CA)

∠BA'C' = ∠BAC (Corresponding angles)

By AAA axiom, ΔBA'C' ~ ΔBAC

Therefore, corresponding sides are proportional,

BC'/BC = BA'/BA = C'A'/CA = 3/4