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Lesson 2-6
#1-17all
Geometric Proof
Short: 1-17 odd
To write a geometric proof, start with the hypothesis
of a conditional.
Apply deductive reasoning.
Hypothesis
Deductive Reasoning
• Definitions • Properties
• Postulates • Theorems
Prove that the conclusion of the conditional is true.
Conclusion
Conditional: If BD is the angle bisector of ABC, and
ABD  1, then DBC  1.
Given: BD is the angle bisector of ABC, and ABD  1.
Prove: DBC  1
Proof:
1. BD is the angle bisector of ABC.
1. Given
2. ABD  DBC
2. Def. of  bisector
3. ABD  1
3. Given
4. DBC  1
4. Transitive Prop. of 
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1.Given:
N is the midpoint of MP , Q is the
midpoint of RP , and PQ  NM .
Prove:
PN  QR
Write a justification for each step.
Proof:
1. N is the midpoint of MP .
1. _________________________________
2. Q is the midpoint of RP .
2. _________________________________
3. PN  NM
3. _________________________________
4. PQ  NM
4. _________________________________
5. PN  PQ
5. _________________________________
6. PQ  QR
6. _________________________________
7. PN  QR
7. _________________________________
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Geometric Proof continued
A theorem is any statement that you can prove. You can use two-column proofs
and deductive reasoning to prove theorems.
Congruent Supplements
Theorem
If two angles are supplementary to the same angle (or to two
congruent angles), then the two angles are congruent.
Right Angle Congruence
Theorem
All right angles are congruent.
Here is a two-column proof of one case of the Congruent Supplements Theorem.
Given:
4 and 5 are supplementary and
5 and 6 are supplementary.
Prove:
4  6
Proof:
Statements
Reasons
1. 4 and 5 are supplementary.
1. Given
2. 5 and 6 are supplementary.
2. Given
3. m4  m5  180
3. Definition of supplementary angles
4. m5  m6  180
4. Definition of supplementary angles
5. m4  m5  m5  m6
5. Substitution Property of Equality
6. m4  m6
6. Subtraction Property of Equality
7. 4  6
7. Definition of congruent angles
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Fill in the blanks to complete the two-column proof
of the Right Angle Congruence Theorem.
2. Given: 1 and 2 are right angles.
Prove: 1  2
Proof:
Statements
Reasons
1. a. ____________________________
1. Given
2. m1  90
2. b. _______________________________
3. c. ____________________________
3. Definition of right angle
4. m1  m2
4. d. _______________________________
5. e. ____________________________
5. Definition of congruent angles
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Problem Solving
1. Given
3. Def. of midpoint
2. Given
4. Given
5. Transitive Prop. of 
6. Def. of midpoint
7. Transitive Prop. of 
Statements
Reasons
1.
Statements
Reasons
1. 1 and 3 are
supplementary. 2 and
4 are supplementary.
1. Given
2. m1  m3  180°
m2  m4  180°
2. Def. of supp. s
3. m1  m3  m2 
m4
3. Subst. Prop. of 
1. a. 1 and 2 are
right angles.
1. Given
2. m1  90
2. b. Definition of right
angle
4. 3  4
4. Given
5. m3  m4
5. Def. of  s
3. c. m2  90°
3. Definition of right
angle
6. m1  m4  m2 
m4
6. Subst. Prop. of 
4. m1  m2
4. d. Transitive Property
of Equality
7. m4  m4
7. Reflex. Prop. of

5. e. 1  2
5. Definition of congruent
angles
8. m1  m2
8. Subtr. Prop. of 
9. 1  2
9. Def. of  s
Challenge
1. Given: 1  4 and ABC is a right .
Possible answer: Prove 2  3.
2. Given: EF  EJ and FG  JH. Possible
answer: Prove EH  EG .
3.
Statements
Reasons
1. KLM and NML are right
angles.
1. Given
2. KLM  NML
2. Rt.   Thm.
3. mKLM  mNML
3. Def. of  s
4. mKLM  m1  m2,
mNML  m3  m4
4.  Add. Post.
5. m1  m2  m3  m4
5. Subst. Prop. of 
6. 2  3
6. Given
7. m2  m3
7. Def. of  s
8. m1  m2  m2  m4
8. Subst. Prop. of 
9. m2  m2
9. Reflex. Prop. of 
10. m1  m4
10. Subtr. Prop. of 
11. 1  4
11. Def. of  s
2. 1  4
3. 3  5
Reading Strategies
1. FD bisects EFC, FC bisects DFB
2. All three angles should be marked
congruent to each other.
3. EFD  CFB
4. Statements and Reasons
5. Transitive Property of Equality
FLOWCHART AND PARAGRAPH
PROOFS
Practice A
1.
Statements
Reasons
1. mBAC  mEAF, mCAD
 mDAE
1. a. Given
2. b. mBAC  mCAD 
mEAF  mDAE
2. Add. Prop.
of 
3. mBAC  mCAD 
mBAD,
mEAF  mDAE  mDAF
3.  Add.
Post.
4. mBAD  mDAF
4. c. Subst.
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