Given a triangle ABC, we are required to construct another triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.

Steps of Construction :

1. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

2. Locate 4 (the greater of 3 and 4 in 3/4 ) points B1, B2, B3 and B4 on BX so that BB1 = B1B = B2B3 = B3B4.

3. Join B4C and draw a line through B3 (the 3rd point, 3 being smaller of 3 and 4 in 3/4) parallel to B4C to intersect BC at C'.

4. Draw a line through C' parallel to the line CA to intersect BA at A' .

Then, △A'BC' is the required triangle.

Let us now see how this construction gives the required triangle.

By Construction 1, BC'/C'C = 3/1.

Therefore, BC/BC' = (BC'+C'C)/BC' = 1+C'C/BC' = 1+(1/3) = 4/3 , i.e., BC' / BC = 3/4

Also C'A' is parallel to CA. Therefore, △ A'BC' ~ △ABC. (Why ?)

So, A'B = AB = A'C'/AC = BC'/BC = 3/4.