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Construct any ∆ABC. Construct ∆ A’BC’ such that AB : A’B = 5:3 and ∆ ABC ~ ∆ A’BC’. 

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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.

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