Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are times the corresponding sides of the isosceles triangle.

Steps of Construction:

Step I: BC = 5 cm is drawn.

Step II: Draw perpendicular bisector of BC, Let the bisector intersect BC at O.

Taking point D as center, and 4 cm radius, draw an arc interesting ⊥ bisector at point A.

Join AB and AC.

Step III: A ray BX is drawn making an acute angle with BC opposite to vertex A.

Step IV: 3 points B₁ , B₂ and B₃ is marked BX.

Step V: B₂ is joined with C to form B₂C as 2 point is smaller. B₃C' is drawn parallel to B₂C as 3 point is greater.

Step VI: C'A' is drawn parallel to CA.

Thus, A'BC' is the required triangle formed.

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.