given: p is equidistant from the endpoints of $\overline{ab}$.
prove: p lies on the perpendicular bisector of $\overline{ab}$.

complete the missing parts of the paragraph proof.
draw a perpendicular from p to $\overline{ab}$. label the intersection c. we are given that $pa = pb$, so $\overline{pa} \cong \overline{pb}$ by the definition of $\boxed{}$. we know that angles pca and pcb are right angles by the definition of $\boxed{}$. $\overline{pc} \cong \overline{pc}$ by the $\boxed{}$. so, triangle acp is congruent to triangle bcp by hl, and $\overline{ac} \cong \overline{bc}$ by $\boxed{}$. since $\overline{pc}$ is perpendicular to and bisects $\overline{ab}$, p is on the perpendicular bisector of $\overline{ab}$ by the definition of perpendicular bisector.

Question

given: p is equidistant from the endpoints of $\overline{ab}$.
prove: p lies on the perpendicular bisector of $\overline{ab}$.

complete the missing parts of the paragraph proof.
draw a perpendicular from p to $\overline{ab}$. label the intersection c. we are given that $pa = pb$, so $\overline{pa} \cong \overline{pb}$ by the definition of $\boxed{}$. we know that angles pca and pcb are right angles by the definition of $\boxed{}$. $\overline{pc} \cong \overline{pc}$ by the $\boxed{}$. so, triangle acp is congruent to triangle bcp by hl, and $\overline{ac} \cong \overline{bc}$ by $\boxed{}$. since $\overline{pc}$ is perpendicular to and bisects $\overline{ab}$, p is on the perpendicular bisector of $\overline{ab}$ by the definition of perpendicular bisector.

Accepted Answer

# Explanation:
## Step1: Identify Congruence for $ \overline{PA} \cong \overline{PB} $
The definition for when two segments are congruent is that they have equal length. Since $ PA = PB $, by the definition of congruent segments, $ \overline{PA} \cong \overline{PB} $.
## Step2: Identify Right Angles
When we draw a perpendicular from $ P $ to $ \overline{AB} $ (labeled as $ \overline{PC} $), the angles $ \angle PCA $ and $ \angle PCB $ are right angles by the definition of perpendicular lines (perpendicular lines form right angles).
## Step3: Identify Common Side $ \overline{PC} $
For triangles $ \triangle ACP $ and $ \triangle BCP $, the side $ \overline{PC} $ is common to both, so $ \overline{PC} \cong \overline{PC} $ by the Reflexive Property of Congruence (a segment is congruent to itself).
## Step4: Identify Congruence of $ \overline{AC} $ and $ \overline{BC} $
Since $ \triangle ACP \cong \triangle BCP $ by HL (Hypotenuse - Leg) Congruence (we have a right angle, hypotenuse $ \overline{PA} \cong \overline{PB} $, and leg $ \overline{PC} \cong \overline{PC} $), the corresponding parts $ \overline{AC} $ and $ \overline{BC} $ are congruent by CPCTC (Corresponding Parts of Congruent Triangles are Congruent). So $ \overline{AC} \cong \overline{BC} $ means $ AC = BC $, so $ \overline{PC} $ bisects $ \overline{AB} $. Also, since $ \overline{PC} $ is perpendicular to $ \overline{AB} $ (from the perpendicular drawn), $ \overline{PC} $ is the perpendicular bisector.

# Answer:
1. For $ \overline{PA} \cong \overline{PB} $: congruent segments
2. For right angles: perpendicular lines
3. For $ \overline{PC} \cong \overline{PC} $: Reflexive Property of Congruence
4. For $ \overline{AC} \cong \overline{BC} $: CPCTC (Corresponding Parts of Congruent Triangles are Congruent)

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QUESTION IMAGE

Question

Explanation:

Step1: Identify Congruence for \( \overline{PA} \cong \overline{PB} \)

Step2: Identify Right Angles

Step3: Identify Common Side \( \overline{PC} \)

Step4: Identify Congruence of \( \overline{AC} \) and \( \overline{BC} \)

Answer: