Draw a triangle ABC with side BC=6cm,AB=5cm & angle ABC=60' . Then construct a triangle with sides 3/4 of the corresponding sides of the triangle ABC.

 A ΔA'BC' whose sides are of the corresponding sides of ΔABC can be drawn as follows.

Step 1

Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.

Step 2

Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

Step 3

Locate 4 points (as 4 is greater in 3 and 4), B₁, B₂, B₃, B₄, on line segment BX.

Step 4

Join B₄C and draw a line through B₃, parallel to B₄C intersecting BC at C'.

Step 5

Draw a line through C' parallel to AC intersecting AB at A'. ΔA'BC' is the required triangle.