Construct a triangle similar to a given such that each of its sides is of the corresponding sides of . It is given that BC = 6 cm, and .

The steps involved in the required construction are:

1) Draw a line segment BC=6 cm.

2) Using a protractor, draw ∠CBD=50° and ∠BCE=60°. BD and CE intersect at point A.

3) Draw any line segment BF, making an acute angle with BC and opposite to the vertex A. Taking B as the center and any radius, draw an arc, intersecting BF at G. Taking G as the center and radius BG, draw an arc, intersecting BF at H. Taking H as the center and radius BG, draw an arc, intersecting BF at I. Join CI.

4) Taking I as the center and any radius, draw an arc., intersecting BF and CI at J and K respectively. Taking H as the center and radius IJ, draw an arc., intersecting BF at L. Taking L as the center and radius JK, draw an arc, intersecting previous arc at M. Join and extend HM, intersecting BC at N.

5) Taking C as the center and any radius, draw an arc., intersecting BC and CA at P and Q respectively. Taking N as the center and radius CP, draw an arc., intersecting BC at R. Taking R as the center and radius PQ, draw an arc, intersecting previous arc at S. Join and extend NS, intersecting AB at O.

6) ∆BNO is the required triangle.