Construct a triangle similar to a given triangle ABC with its sides equal to 3/4 of the corresponding sides of the triangle ABC (i.e. of scale factor 3/4).
Given scale factor = 3/4
Steps of Construction:
- Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
- Locate 4 (the greater of 3 and 4 in 3/4 ) points B_1,B_2,B_3 and B_4 on BX so that 〖BB〗_1=B_1 B_2=B_2 B_3=B_3 B_4.
- Join B_4C and draw a line through B_3 (the 3rd point, 3 being smaller of 3 and 4 in 3/4) parallel to B_4 C to intersect BC at C′.
- Draw a line through C′ parallel to the line CA to intersect BA at A′.
Thus, Δ A′BC′ is the required triangle
Also, A’C’ is parallel to AC
So, the will make the same angle with line BC
∴ ∠ A’C’B = ∠ ACB
In Δ A’BC’ and ABC
∠ B = ∠ B
∠ A’C’B = ∠ ACB
Δ A’BC’ ∼ Δ ABC
Since corresponding sides of
similar triangles are in the same ratio
(A^′ B)/AB=(A^′ C^′)/AC=(BC^′)/BC
So, (A^′ B)/AB=(A^′ C^′)/AC=(BC^′)/BC =3/4.
This justifies the construction.