Construct a triangle ABC with side BC = 7 cm, B = 45^o, A = 105^o. Then construct another triangle whose sides are times the corresponding sides of the ABC.

Steps of constructions:

1. Draw a line segment BC = 7 cm, and draw ∠YBC = 45°.

2. Draw ∠ZCB = 30° ; and let CZ and BY intersect at A. Therefore, ΔABC is ready.

[ ∠C = 30°, as ∠A = 105° ; ∠B = 45° and we know, By angle sum property of triangle,

∠A + ∠B + ∠C = 180°

⇒ 45° + 105° + ∠C = 180°

⇒ ∠C = 30° ]

3. Below BC, make an acute angle ∠CBX.

4. Along BX mark 4 (greater of 3 and 4 in 3/4) points B1, B2, B3,B4 at equal distances.

5. Join B4C

6. From point B3, make a line B3C' parallel to B4C , meeting BC at C' [Choosing B3 because we need a 3:4 ratio]

7. From point B draw a line C'A' parallel to CA.

8.A'BC' is the triangle, whose sides are three - fourth of ∆ABC