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 $overline{pa}congoverline{pb}$ by the definition of. we know that angles pca and pcb are right angles by the definition of. $overline{pc}congoverline{pc}$ by the. so, triangle acp is congruent to triangle bcp by hl, and $overline{ac}congoverline{bc}$ by. 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.

Explanation:

Step1: Define congruent segments

Since $PA = PB$, by the definition of congruent segments, $\overline{PA}\cong\overline{PB}$.

Step2: Define perpendicular lines

We draw a perpendicular from $P$ to $\overline{AB}$ at $C$. By the definition of perpendicular lines, $\angle PCA$ and $\angle PCB$ are right - angles.

Step3: Identify reflexive property

$\overline{PC}\cong\overline{PC}$ by the reflexive property of congruence.

Step4: Apply HL congruence theorem

Since $\overline{PA}\cong\overline{PB}$ (hypotenuses) and $\overline{PC}\cong\overline{PC}$ (legs), $\triangle ACP\cong\triangle BCP$ by the HL (Hypotenuse - Leg) congruence theorem.

Step5: Use CPCTC

Since $\triangle ACP\cong\triangle BCP$, by CPCTC (Corresponding Parts of Congruent Triangles are Congruent), $\overline{AC}\cong\overline{BC}$.

Answer:

1. Congruent segments
2. Perpendicular lines
3. Reflexive property of congruence
4. CPCTC