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Example 2
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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

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Steps
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of Construction
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:
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- 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.

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Justification
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Here,

(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

Now,

In Δ A’BC’ and ABC

∠ B = ∠ B

∠ A’C’B = ∠ ACB

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Δ
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A’BC’ ∼
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Δ
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ABC
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Since corresponding sides of

similar triangles are in the same ratio

(A^′ B)/AB=(A^′ C^′)/AC=(BC^′)/BC

So,
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(A^′ B)/AB=(A^′ C^′)/AC=(BC^′)/BC
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=5/3
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**
.
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This justifies the construction.