To determine the similarity criterion that proves triangles ABC and A'B'C' are similar, we first need to analyze the provided coordinates of the vertices of both triangles.
Given Coordinates:
-
Triangle ABC:
- A(-3, -2)
- B(4, -2)
- C(-3, 3)
-
Triangle A'B'C':
- A'(-6, -2)
- B'(8, -2)
- C'(-6, 8)
Step 1: Calculate the lengths of the sides of both triangles.
Triangle ABC:
-
Length of AB: \[ AB = \sqrt{(4 - (-3))^2 + (-2 - (-2))^2} = \sqrt{(4 + 3)^2 + (0)^2} = \sqrt{7^2} = 7 \]
-
Length of BC: \[ BC = \sqrt{((-3) - 4)^2 + (3 - (-2))^2} = \sqrt{(-7)^2 + (5)^2} = \sqrt{49 + 25} = \sqrt{74} \]
-
Length of CA: \[ CA = \sqrt{((-3) - (-3))^2 + (3 - (-2))^2} = \sqrt{(0)^2 + (5)^2} = 5 \]
Triangle A'B'C':
-
Length of A'B': \[ A'B' = \sqrt{(8 - (-6))^2 + (-2 - (-2))^2} = \sqrt{(8 + 6)^2 + (0)^2} = \sqrt{14^2} = 14 \]
-
Length of B'C': \[ B'C' = \sqrt{((-6) - 8)^2 + (8 - (-2))^2} = \sqrt{(-14)^2 + (10)^2} = \sqrt{196 + 100} = \sqrt{296} \]
-
Length of C'A': \[ C'A' = \sqrt{((-6) - (-6))^2 + (8 - (-2))^2} = \sqrt{(0)^2 + (10)^2} = 10 \]
Step 2: Determine if the triangles are similar by looking at the ratio of the sides.
Ratios of corresponding sides:
-
Ratio of AB to A'B': \[ \text{Ratio} = \frac{AB}{A'B'} = \frac{7}{14} = \frac{1}{2} \]
-
Ratio of BC to B'C': \[ \text{Ratio} = \frac{BC}{B'C'} = \frac{\sqrt{74}}{\sqrt{296}} = \frac{\sqrt{74}}{2\sqrt{74}} = \frac{1}{2} \]
-
Ratio of CA to C'A': \[ \text{Ratio} = \frac{CA}{C'A'} = \frac{5}{10} = \frac{1}{2} \]
Conclusion:
Since all pairs of corresponding sides of triangles ABC and A'B'C' are in proportion (each has a ratio of \(\frac{1}{2}\)), we can conclude that triangles ABC and A'B'C' are similar by the Side-Side-Side (SSS) similarity criterion.
Final Answer:
The similarity criterion that proves the two triangles are similar is SSS~.
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