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), B_{1}, B_{2}, B_{3}, B_{4}, on line segment BX.

**Step 4**

Join B_{4}C and draw a line through B_{3}, parallel to B_{4}C intersecting BC at C'.

**Step 5**

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