Which similarity criterion proves the two triangles above are similar?

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:

1. Length of AB: \[ AB = \sqrt{(4 - (-3))^2 + (-2 - (-2))^2} = \sqrt{(4 + 3)^2 + (0)^2} = \sqrt{7^2} = 7 \]

2. Length of BC: \[ BC = \sqrt{((-3) - 4)^2 + (3 - (-2))^2} = \sqrt{(-7)^2 + (5)^2} = \sqrt{49 + 25} = \sqrt{74} \]

3. Length of CA: \[ CA = \sqrt{((-3) - (-3))^2 + (3 - (-2))^2} = \sqrt{(0)^2 + (5)^2} = 5 \]

Triangle A'B'C':

1. Length of A'B': \[ A'B' = \sqrt{(8 - (-6))^2 + (-2 - (-2))^2} = \sqrt{(8 + 6)^2 + (0)^2} = \sqrt{14^2} = 14 \]

2. Length of B'C': \[ B'C' = \sqrt{((-6) - 8)^2 + (8 - (-2))^2} = \sqrt{(-14)^2 + (10)^2} = \sqrt{196 + 100} = \sqrt{296} \]

3. 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:

1. Ratio of AB to A'B': \[ \text{Ratio} = \frac{AB}{A'B'} = \frac{7}{14} = \frac{1}{2} \]

2. Ratio of BC to B'C': \[ \text{Ratio} = \frac{BC}{B'C'} = \frac{\sqrt{74}}{\sqrt{296}} = \frac{\sqrt{74}}{2\sqrt{74}} = \frac{1}{2} \]

3. 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~.