Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are $\frac{2}{3}$ of the corresponding sides of the first triangle.

Steps of constructions:

1. Draw BC = 6 cm. With B and C as centres and radii, 5 cm and 4 cm respectively draw arcs to intersect at A. ΔABC is obtained.
2. Draw ray BX making an acute angle with BC.
3. Mark 3 (since, 3 > 2 in the ratio 2/3) points B₁, B₂, B₃ on BX such that BB₁ = B₁B₂ = B₂B₃.
4. Join B₃C and draw a line through B₂ (second point where 2 < 3 in the ratio) parallel to B₃C meeting BC at C'.
5. Draw a line thorough C' parallel to CA to meet BA at A’. Now ΔA'BC' is the required triangle similar to ΔABC where BC'/BC = BA'/BA = C'A'/CA = 2/3

Proof:

In ΔBB₃C, B₂C' is parallel to B₃C.

Hence by Basic proportionality theorem,

B₂B₃/BB₂ = C'C/BC' = 1/2

Adding 1 to both the sides of C'C/BC' = 1/2,

C'C/BC' + 1 = 1/2 + 1

(C'C + BC')/BC' = 3/2

BC/BC' = 3/2

or BC'/BC = 2/3 ....(1)

Consider ΔBA'C' and ΔBAC

∠A'BC' = ∠ABC (Common)

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

Hence by AA similarity, ΔBA'C' ~ ΔBAC

Corresponding sides are proportional

BA'/BA = C'A'/CA = BC'/BC = 2/3 [From equation(1)]