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**GoldendoodleMom** Complete this proof using axioms and already proven theorems.

Given: Point P is equidistant from endpoints X and Y of line XY..

Prove: P is on the perpendicular bisector of XY

Proof:

Case 1: P is on XY. By the Given, P is the midpoint of XY so it is on the perpendicular bisector.

Case 2:

1. Draw PX and PY. (On 2 points there is exactly 1 line)

2. Let M be the midpoint of XY. (Midpoint Thm)

3. Draw PM (On 2 points there is exactly 1 line)

4. PX = PY (Given)

5. XM = YM (Definition of midpoint)

6. PM = PM (Reflexive Property of equality)

7. $\displaystyle \triangle XPM \cong \triangle YPM$ (SSS Postulate)

8. $\displaystyle \angle XPM \cong \angle YPM $ and $\displaystyle m\angle XPM = m\angle YPM$ (CPCTC and definition of congruency)

9. $\displaystyle \angle XPM \ \ and \ \ \angle YPM$ make up a Linear Pair (Definition of Linear Pair)

10. $\displaystyle m\angle XPM + m\angle YPM = 180$ (If two angles form a linear pair then they are supplementary)

11. $\displaystyle m\angle XPM + m\angle XPM = 180$ (Substitution using #8 and #10)

12. $\displaystyle 2 m\angle XPM = 180$ (Addition)

13. $\displaystyle m\angle XPM = 90$ (Division)

Similarly, you can show that $\displaystyle m\angle YPM = 90$

14.$\displaystyle \angle XPM$ is a right angle. (Definition of right angle)

15. $\displaystyle \overline{PM} \perp \overline {XY}$ (If two lines meet to form right angles, then they are perpendicular)

16. P lies on $\displaystyle \overline{PM}$ (Step #3)

Q.E.D. P lies on the perpendicular bisector of $\displaystyle \overline {XY}$

I'm stuck after that. Any help would be greatly appreciated.