Construct a triangle ABC with sides BC = 7 cm, ∠B = 45° and ∠A = 105°. Then construct a triangle whose sides are `3/4` times the corresponding sides of ∆ABC.
Steps of construction:
1. Draw line BC = 7 cm
2. At B, construct ∠CBY = 45° and at C, construct 180° −(105° + 45°) = 30°.
3. Mark the point of intersection of ∠B = 45° and ∠C = 30° as A. Thus, ∆ABC is obtained.
4. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
5. Locate 4 (3 < 4) points B1, B2, B3 and B4 on BX such that BB1 = B1B2 = B2B3 = B3B4.
6. Join B4C and draw a line through B3 parallel to B4C to intersect BC at C'.
7. Draw a line through C′ parallel to the line CA to intersect BA at A′
Then, ΔA′BC′ is the required triangle similar to the ΔABC