Given points $$ P(-5,-1), R(-3,1), O(1,5), V(4,-7) $$, and $$ E(-2,-3) $$, which of the following proves that $$ riangle P O V \sim riangle P R E $$ ? By the Distance Formula, $$ P O=6 \sqrt{2}, P R=2 \sqrt{2}, P V=3 \sqrt{13} $$, and $$ P E= $$ Since, $$ \frac{P O}{P R}=\frac{P V}{P E}=\frac{3}{1}=3 $$, and $$ \angle P \cong \angle P $$ by the Reflexive Property, $$ riangle P O V \sim riangle P R E $$ by Bi By the Distance Formula, $$ P O=2 \sqrt{2}, P R=6 \sqrt{2}, P V=\sqrt{13} $$, and $$ P E=3 \downarrow $$ Since, $$ \frac{P O}{P R}=\frac{P V}{P E}=\frac{3}{1}=3 $$, and $$ \angle P \cong \angle P $$ by the Reflexive Property, $$ riangle P O V \sim riangle P R E $$ by By the Distance Formula, $$ P O=2 \sqrt{2}, P R=6 \sqrt{2}, P V=\sqrt{13} $$, and $$ P E=3 \downarrow $$ Since, $$ \frac{P O}{P R}=\frac{P V}{P E}=\frac{3}{1}=3 $$, and $$ \angle P \cong \angle P $$ by the Reflexive Property, $$ riangle P O V \sim riangle P R E $$ by S. By the Distance Formula, $$ P O=6 \sqrt{2}, P R=2 \sqrt{2}, P V=3 \sqrt{13} $$, and $$ P E= $$ Since, $$ \frac{P O}{P R}=\frac{P V}{P E}=\frac{3}{1}=3 $$, and $$ \angle P \cong \angle P $$ by the Reflexive Property, $$ riangle P O V \sim riangle P R E $$ by S.

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Given points $$ P(-5,-1), R(-3,1), O(1,5), V(4,-7) $$, and $$ E(-2,-3) $$, which of the following proves that $$ 	riangle P O V \sim 	riangle P R E $$ ? By the Distance Formula, $$ P O=6 \sqrt{2}, P R=2 \sqrt{2}, P V=3 \sqrt{13} $$, and $$ P E= $$ Since, $$ \frac{P O}{P R}=\frac{P V}{P E}=\frac{3}{1}=3 $$, and $$ \angle P \cong \angle P $$ by the Reflexive Property, $$ 	riangle P O V \sim 	riangle P R E $$ by Bi By the Distance Formula, $$ P O=2 \sqrt{2}, P R=6 \sqrt{2}, P V=\sqrt{13} $$, and $$ P E=3 \downarrow $$ Since, $$ \frac{P O}{P R}=\frac{P V}{P E}=\frac{3}{1}=3 $$, and $$ \angle P \cong \angle P $$ by the Reflexive Property, $$ 	riangle P O V \sim 	riangle P R E $$ by By the Distance Formula, $$ P O=2 \sqrt{2}, P R=6 \sqrt{2}, P V=\sqrt{13} $$, and $$ P E=3 \downarrow $$ Since, $$ \frac{P O}{P R}=\frac{P V}{P E}=\frac{3}{1}=3 $$, and $$ \angle P \cong \angle P $$ by the Reflexive Property, $$ 	riangle P O V \sim 	riangle P R E $$ by S. By the Distance Formula, $$ P O=6 \sqrt{2}, P R=2 \sqrt{2}, P V=3 \sqrt{13} $$, and $$ P E= $$ Since, $$ \frac{P O}{P R}=\frac{P V}{P E}=\frac{3}{1}=3 $$, and $$ \angle P \cong \angle P $$ by the Reflexive Property, $$ 	riangle P O V \sim 	riangle P R E $$ by S.

UpStudy AI · Accepted Answer

To prove that $$\triangle P O V \sim \triangle P R E$$, we can use the **Side-Angle-Side (SAS) Similarity Criterion**. Here's how:

1. **Calculate the Distances Using the Distance Formula:**

- **Distance $$ P O $$:**
     $$
     P(-5, -1) \text{ to } O(1, 5) = \sqrt{(1 - (-5))^2 + (5 - (-1))^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}
     $$
   
   - **Distance $$ P R $$:**
     $$
     P(-5, -1) \text{ to } R(-3, 1) = \sqrt{(-3 - (-5))^2 + (1 - (-1))^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}
     $$
   
   - **Distance $$ P V $$:**
     $$
     P(-5, -1) \text{ to } V(4, -7) = \sqrt{(4 - (-5))^2 + (-7 - (-1))^2} = \sqrt{9^2 + (-6)^2} = \sqrt{117} = 3\sqrt{13}
     $$
   
   - **Distance $$ P E $$:**
     $$
     P(-5, -1) \text{ to } E(-2, -3) = \sqrt{(-2 - (-5))^2 + (-3 - (-1))^2} = \sqrt{3^2 + (-2)^2} = \sqrt{13}
     $$

2. **Compare the Ratios of Corresponding Sides:**
   
   $$
   \frac{P O}{P R} = \frac{6\sqrt{2}}{2\sqrt{2}} = 3
   $$
   
   $$
   \frac{P V}{P E} = \frac{3\sqrt{13}}{\sqrt{13}} = 3
   $$
   
   Since $$\frac{P O}{P R} = \frac{P V}{P E}$$, the ratios of the corresponding sides are equal.

3. **Verify the Included Angle:**
   
   Both triangles share $$\angle P$$, which is congruent by the **Reflexive Property**.

4. **Apply the SAS Similarity Criterion:**
   
   Since two pairs of corresponding sides are in proportion and the included angles are congruent, by **SAS Similarity**, the triangles are similar.

**Conclusion:**

$$
\triangle P O V \sim \triangle P R E \quad \text{by SAS Similarity Criterion}
$$

**Final Statement:**

By correctly calculating the side lengths (PO = 6√2, PR = 2√2, PV = 3√13, PE = √13), showing PO/PR = PV/PE = 3, and noting that ∠P is congruent by the Reflexive Property, we conclude that △POV ~ △PRE by SAS similarity.
To prove that $$\triangle P O V \sim \triangle P R E$$, we use the Side-Angle-Side (SAS) Similarity Criterion. By calculating the distances and showing that the ratios of corresponding sides are equal with the included angles congruent, we conclude that the triangles are similar by SAS.

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