Analysis:
As shown in the figure,
Let B – A’ – A and B – C’ -C
∆ ABC – A’BC’ … [Given]
∴ ∠ABC ≅ ∠A’BC’ …[Corresponding angles of similar trianglesi
∴ Sides of ∆ABC are longer than corresponding sides of ∆A’BC’. Rough figure
∴ the length of side BC’ will be equal to 3 parts out of 5 equal parts of side BC.
So if we construct ∆ABC, point C’ will be on side BC, at a distance equal to 3 parts from B.
Now A’ is the point of intersection of AB and a line through C’ , parallel to CA.
Let ∆ABC be any triangle constructed such that AB = 7 cm, BC = 7 cm and AC = 6 cm.