Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are (2/5) of the corresponding sides of it.

Steps of Construction:

Step I: AB = 6 cm is drawn.

Step II: With A as a centre and radius equal to 4cm, an arc is draw. Again, with B as a centre and radius equal to 5 cm an arc is drawn on same side of AB intersecting previous arc at C.

Step III: AC and BC are joined to form ΔABC.

Step IV: A ray AX is drawn making an acute angle with AB below it.

Step V: 5 equal points (sum of the ratio = 2 + 3 =5) is marked on AX as A₁ A₂....A₅

Step VI: A₅B is joined. A₂ B' is drawn parallel to A5B and B'C' is drawn parallel to BC.

ΔAB'C' is the required triangle

Justification:

∠A(Common)

∠C = ∠C' and ∠B = ∠ B' (corresponding angles)

Thus ΔAB'C' ~ ΔABC by AAA similarity condition

From the figure,

Hence Proved.