Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are $\frac{7}{5}$ of the corresponding sides of the first triangle.

Steps of construction:

1. Draw BC = 7cm with B and C as centres and radii 5 cm and 6 cm respectively. Draw arcs to intersect at A. ΔABC is obtained.
2. Draw ray BX making ∠CBX an acute angle.
3. Mark 7 points (greater of 7 and 5 in 7/5 ) B₁, B₂,………B₇ on BX such that BB₁ = B₁B₂ =............... = B₆B₇
4. Join B₅ (smaller of 7 and 5 in 7/5 which is the 5^th point) to C and draw B₇C' parallel to B₅C intersecting the extension of BC at C'.
5. Through C' draw C'A' parallel to CA to meet the extension of BA at A’. Now, ΔA' B'C' is the required triangle similar to ΔABC where BA'/BA = C'A'/CA = BC'/BC = 7/5

Proof:

In ΔBB₇C', B₃C is parallel to B₇C'

Hence by Basic proportionality theorem,

B₅B₇/BB₅ = CC'/BC = 2/5

Adding 1 to both the sides of CC'/BC = 2/5,

CC'/BC + 1 = 2/5 + 1

(BC + CC')/BC = 7/5

BC'/BC = 7/5

Consider ΔBAC and ΔBA'C'

∠ABC = ∠A'BC' (Common)

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

By AA criteria, ΔBAC ∼ ΔBA'C'

∴ Corresponding sides are proportional

Hence,

BA'/BA = C'A'/A = BC'/BC = 7/5