Question

Which similarity criterion proves the two triangles above are similar?

(1 point)
Responses
ABC (-3, -2), (4, -2), (-3,3) to A' B' C' (-6, -2), (8, -2), (-6,8) This is a dilation enlargement Explain your reasoning by proving the answer above and show your work. You may type it out or upload a picture. You must fully explain and show all work in order to receive full credit.
AA~
AA~

SSS~
SSS~

SAS~
SAS~

AAS~
AAS~
Question 2

Answer 1

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~**.