Draw a triangle ABC in which AB=5 cm, BC=6 cm and ∠ABC= 60°. Construct another triangle similar to ΔABC with scale factor .

Steps of construction:

Step 1:

Construct a line BC of 6 cm.

Step 2:

Draw angle of 60° at B.

Step 3:

From B draw an arc of 5 cm. The point intersects at A. Join AC.

Step 4:

Draw BX.

Step 5:

Mark the greater of i.e. 7 arcs in equal distance on BX.

Step 6:

Join B₇C and then draw a line through B₅ parallel to B₇C.

Step 7:

From C₁ draw a line parallel to AC.

Thus, A₁BC₁ is the required triangle.

Justification:

Since the scale factor is ,

We need to prove,

By construction,

… (1)

Also, A₁C₁ is parallel to AC.

So, this will make same angle with BC.

∴ ∠A₁C₁B = ∠ACB …. (2)

Now,

In ΔA₁BC₁ and ΔABC

∠ B = ∠ B (common)

∠A₁C₁B = ∠ACB (from 2)

ΔA₁BC₁∼ ΔABC

Since corresponding sides of similar triangles are in same ratio.

From (1)

Hence construction is justified.