Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are $\frac{4}{3}$ times the corresponding sides of ∆ ABC.

Steps of construction:

1. Draw BC = 7cm and at B, make an angle ∠CBY = 45° and at C, make ∠BCZ = 30° [Since, 180° - (45° + 105° )]. Both BY and CZ intersect at A and thus ΔABC is constructed.
2. Draw the ray BX so that ∠CBX is acute.
3. Mark 4 (since 4 > 3 in 4/3) points B₁, B₂, B₃, B₄ on BX such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄
4. Join B₃ third point on BX, (since 3 < 4 in 4/3) to C and draw B C' parallel to BC such that C’ lies on the extension of BC.
5. Draw C’A’ parallel to CA to intersect the extension of BA at A’. Now, triangle A'BC' is the required triangle similar to ΔABC where, BA'/BA = BC'/BC = C'A'/CA = 4/3

Proof:

In ΔBB₄C', B₃C || B₄C'

Hence by Basic proportionality theorem,

B₃B₄/BB₃ = CC'/BC = 1/3

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

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

BC'/BC = 4/3

Consider ΔBA'C' and ΔBAC

∠A'BC' = ∠ABC = 45°

∠BC'A' = ∠BCA = 30° (Corresponding angles as CA || C'A')

∠BA'C' = ∠BAC = 105° (Corresponding angles)

By AAA axiom, ΔBA'C' ~ ΔBAC

Hence corresponding sides are proportional

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