# Complete this geometry proof? Been trying ALL night!?

Converse of the Perpendicular Bisector Theorem: Any point equidistant from the endpoints of the segment lies on the perpendicular bisector of the segment. I'm so confused, any help would be great!

Given: AD = BD

AE = BE

Prove: DE is the perpendicular bisector of AB

Statements

1. AD = BD

AE = BE

2. DE = DE

3. triangle ADE @ triangle BDE

4. Angle AED is congruent to Angle BED

5. CE is congruent to CE

6. AEC is cong. to BEC

7. Angle ACE is cong. to BCE

8. Angle ACE and Angle BCE linear pair

9. Angle ACE plus Angle BCE =180

10. AB is perp. to DE Angle ACE =90 Angle BCE = 90

11.AC is cong. to BC

12. Angle ACE and BCE right angles

13. AB is perp. to DE

14. C is the midpoint of AB

15. DE is perp. bisector

I need all the reasons

### 3 Answers

- ?Lv 49 years agoFavorite Answer
There should be a picture to go with this, right?

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- hayharbrLv 79 years ago
Without seeing the picture I can't be sure of this but here is my guess:

Statements

1. AD = BD Given

AE = BE

2. DE = DE Reflexive

3. triangle ADE @ triangle BDE SSS

4. Angle AED is congruent to Angle BED Corresponding parts of ≅ triangles are ≅

5. CE is congruent to CE Reflexive

6. AEC is cong. to BEC SAS

7. Angle ACE is cong. to BCE same as 4

8. Angle ACE and Angle BCE linear pair definition linear pair

9. Angle ACE plus Angle BCE =180 linear pairs are supplementary + def. supplementary

10. AB is perp. to DE Angle ACE =90 Angle BCE = 90 Since they are cong; each is half of 180, and def. perpendicular

11.AC is cong. to BC same as 4

12. Angle ACE and BCE right angles def right angles

13. AB is perp. to DE def perpendicular

14. C is the midpoint of AB def midpoint

15. DE is perp. bisector def perp. bisector

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- 4 years ago
You know that <FED and <DEW are complementary which means that if add their price it might be 90º ... So A. The angles are complementary B. The angles are co linear (they share the segment ED) C. On account that A and B are true then m<FEW=90º D. A right attitude= 90º

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