Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are $\frac{5}{3}$ times the corresponding sides of the given triangle.

Steps of constructions:

1. Draw BC = 4. At B, make an angle ∠CBY = 90° and mark A on BY such that BA = 3 cm. Join A to C. Thus ΔABC is constructed.
2. Draw the ray BX so that ∠CBX is acute.
3. Mark 5 (since, 5 > 3 in 5/3) points B₁, B₂, B₃, B₄, B₅ on BX so that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅
4. Join B₃ (3^rd point on BX as 3 < 5) to C and draw B₅C' parallel to B₃C so that C' lies on the extension of BC.
5. Draw C'A' parallel to CA to intersect of the extension of BA at A’. Now ΔBA'C' is the required triangle similar to ΔBAC where BA'/BA = BC'/BC = C'A'/CA = 5/3

Proof:

In ΔBB₅C', B₃C || B₃C'

Hence by Basic proportionality theorem,

B₃B₅/BB₃ = CC'/BC = 2/3

CC'/BC + 1 = 2/3 + 1 (Adding 1)

(CC' + BC)/BC = 5/3

BC'/BC = 5/3

Consider ΔBAC and ΔBA'C'

∠ABC = ∠A'BC' = 90°

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

∠BAC = ∠BA'C'

By AAA axiom, ΔBAC ~ ΔBA'C'

Therefore, corresponding sides are proportional,

Hence,

BA'/BA = BC'/BC = C'A'/CA = 5/3