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Example 1
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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

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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 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

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Justification
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Here,

BC^′/BC=(BB_3)/(BB_4 )=3/4.

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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=3/4
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.
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This justifies the construction.