Construct a triangle similar to a given triangle ABC with its sides equal its to 5/3 of the corresponding sides of the triangle ABC (scale factor 5/3).
Given scale factor = 5/3
Steps of Construction:
- Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
- Locate 5 (the greater of 5 and 3 in 5/4 ) points B_1,B_2,B_3,B_4 and B_5 on BX so that 〖BB〗_1=B_1 B_2=B_2 B_3=B_3 B_4=B_4 B_5.
- Join B_3 (3rd point as 3 is smaller in 5/3) to C and draw a line through B_5 parallel to B_3 C to intersecting extended BC at C′.
- Draw a line through C′ parallel to line CA to intersect extended BA at A′.
Thus, Δ A′BC′ is the required triangle.
(BC^′)/BC=(BB_5)/(BB_3 ) = 5/3
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 =5/3.
This justifies the construction.