To determine if the two triangles \( \triangle ABC \) and \( \triangle A'B'C' \) are similar, we can use the criterion of proportionality of the lengths of corresponding sides.
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Calculate the lengths of the sides of \( \triangle ABC \):
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Length \( AB \): \[ AB = \sqrt{(4 - (-3))^2 + (-2 - (-2))^2} = \sqrt{(4 + 3)^2 + 0^2} = \sqrt{7^2} = 7 \]
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Length \( BC \): \[ BC = \sqrt{((-3) - 4)^2 + (3 - (-2))^2} = \sqrt{(-7)^2 + 5^2} = \sqrt{49 + 25} = \sqrt{74} \]
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Length \( CA \): \[ CA = \sqrt{((-3) - (-3))^2 + (3 - (-2))^2} = \sqrt{0^2 + (3 + 2)^2} = \sqrt{5^2} = 5 \]
So, the side lengths of \( \triangle ABC \) are:
- \( AB = 7 \)
- \( BC = \sqrt{74} \)
- \( CA = 5 \)
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Calculate the lengths of the sides of \( \triangle A'B'C' \):
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Length \( A'B' \): \[ A'B' = \sqrt{(8 - (-6))^2 + (-2 - (-2))^2} = \sqrt{(8 + 6)^2 + 0^2} = \sqrt{14^2} = 14 \]
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Length \( B'C' \): \[ B'C' = \sqrt{((-6) - 8)^2 + (8 - (-2))^2} = \sqrt{(-14)^2 + 10^2} = \sqrt{196 + 100} = \sqrt{296} \]
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Length \( C'A' \): \[ C'A' = \sqrt{((-6) - (-6))^2 + (8 - (-2))^2} = \sqrt{0^2 + (8 + 2)^2} = \sqrt{10^2} = 10 \]
So, the side lengths of \( \triangle A'B'C' \) are:
- \( A'B' = 14 \)
- \( B'C' = \sqrt{296} \)
- \( C'A' = 10 \)
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Check the proportionality:
Now we will check the ratios of the corresponding sides:
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For \( AB \) and \( A'B' \): \[ \frac{AB}{A'B'} = \frac{7}{14} = \frac{1}{2} \]
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For \( BC \) and \( B'C' \): \[ \frac{BC}{B'C'} = \frac{\sqrt{74}}{\sqrt{296}} = \frac{\sqrt{74}}{\sqrt{4 \times 74}} = \frac{1}{2} \]
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For \( CA \) and \( C'A' \): \[ \frac{CA}{C'A'} = \frac{5}{10} = \frac{1}{2} \]
Since all corresponding sides have the same ratio: \[ \frac{AB}{A'B'} = \frac{BC}{B'C'} = \frac{CA}{C'A'} = \frac{1}{2} \]
This shows that the two triangles \( \triangle ABC \) and \( \triangle A'B'C' \) are similar by the Side-Side-Side (SSS) similarity criterion.
Thus, the similarity criterion that proves \( \triangle ABC \sim \triangle A'B'C' \) is the proportionality of the corresponding sides.