
Given BC = 5.4 cm, ∠B = 45°, ∠A = 105°
Step of construction
1.Draw BC = 5.4 cm
2.Construct ∠XBC = 45° and
∠YCB = 180° – (45° + 105°) = 30° at B and C, respectively intersecting at A.
3. Below BC, make an acute ∠CBZ
4. Along BZ, mark off 4 arcs: B1, B2, B3, B4 such that
BB1 = B1B2 = B2B3 = B3B4
5. Join B3 C
6. From B4, draw B4 D || B3 C, meeting BC produced at D
7. From D, draw DE || AC, meeting BA produced at E.
Then EBD is the required triangle whose sides are 4/3times the corresponding sides of ∆ABC