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.

#### Solution

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 B_{1}, B_{2}, B_{3} and B_{4} on BX such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}.

6. Join B_{4}C and draw a line through B3 parallel to B_{4}C 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