# Geometry Proof

Printable View

• Sep 5th 2008, 04:36 AM
GoldendoodleMom
Geometry Proof
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....

I'm stuck after that. Any help would be greatly appreciated.
• Sep 5th 2008, 05:14 AM
kalagota
Quote:

Originally Posted by 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....

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

i dont know if you have already proven the theorems i will be using..

1. P is not on XY, there is a line passing P that is perpendicular to XY. (Parallel/Perpendicular Postulate)

2. Let M be the point of intersection of XY and the perpendicular passing P. (Non-parallel lines intersect)

3. PM is perpendicular to XY (By 1 and 2)

4. PMX and PMY form right triangles. (Def. of Right triangles)

5. PX is congruent to PY (Given)

6. PM is congruent to itself (Reflexive)

7. triangles PMX and PMY are congruent (Hypotenuse-Leg Theorem)

8. MX is congruent to MY. (CPCTC)

9. M is the midpoint of XY (Mdpt theorem (or def of mdpt of a segment))

10. PM is the perpendicular bisector. (by 3 and 9)

tell me what theorems are not yet proven so that i can revise it for you..
• Sep 5th 2008, 02:41 PM
masters
Quote:

Originally Posted by 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.

Here's another approach. I Don't know how much detail you need. Sometimes, geometry teachers can be pretty picky.