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1STANDARDS 1,2,3END SHOW
PARAGRAPH PROOF 1 PARAGRAPH PROOF 2
PARAGRAPH PROOF 3 PARAGRAPH PROOF 4
Two column proofs may also be written as paragraph proofs. All we need is to the statements and reasons with transition words, phrases or sentences, as may be seen in the following examples that convert two column proofs we did before to paragraph proofs.
“glue”
PARAGRAPH PROOF 5
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2
Standard 1:
Students demonstrate understanding by identifying and giving examples of undefined terms, axioms, theorems, and inductive and deductive reasoning.
Standard 2:
Students write geometric proofs, including proofs by contradiction.
Standard 3:
Students construct and judge the validity of a logical argument and give counterexamples to disprove a statement.
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3
Estándar 1:
Los estudiantes demuestran entendimiento en identificar ejemplos de términos indefinidos, axiomas, teoremas, y razonamientos inductivos y deductivos.
Standard 2:
Los estudiantes escriben pruebas geométricas, incluyendo pruebas por contradicción.
Standard 3:
Los estudiantes construyen y juzgan la validéz de argumentos lógicos y dan contra ejemplos para desaprobar un estatuto.
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4
STANDARDS 1,2,3Review the following two column proof and then convert it to paragraph proof:
Two Column Proof:
Statements Reasons
(1) (1) Given
(2) (2)
(3) (3)
(4) (4)
L is midpoint of KM
KL LM Definition of Midpoint
LM AB Given
KL AB of segments is transitive.
Given:
Prove:
MLK
BA
L is midpoint of KM
LM AB
KL AB
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5
STANDARDS 1,2,3
Two Column Proof:
Statements Reasons
(1) (1) Given
(2) (2)
(3) (3)
(4) (4)
L is midpoint of KM
KL LM Definition of Midpoint
LM AB Given
KL AB of segments is transitive.
Given:
Prove:
MLK
BA
L is midpoint of KM
LM AB
KL AB
The given states that L is midpoint of KM
Convert to PARAGRAPH PROOF:
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
6
STANDARDS 1,2,3
Two Column Proof:
Statements Reasons
(1) (1) Given
(2) (2)
(3) (3)
(4) (4)
L is midpoint of KM
KL LM Definition of Midpoint
LM AB Given
KL AB of segments is transitive.
Given:
Prove:
MLK
BA
L is midpoint of KM
LM AB
KL AB
The given states that L is midpoint of KM and therefore KL LM by definition
of midpoint.
Convert to PARAGRAPH PROOF:
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7
STANDARDS 1,2,3
Two Column Proof:
Statements Reasons
(1) (1) Given
(2) (2)
(3) (3)
(4) (4)
L is midpoint of KM
KL LM Definition of Midpoint
LM AB Given
KL AB of segments is transitive.
Given:
Prove:
MLK
BA
L is midpoint of KM
LM AB
KL AB
The given states that L is midpoint of KM and therefore KL LM by definition
of midpoint. We also have that the given states LM AB
Convert to PARAGRAPH PROOF:
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8
STANDARDS 1,2,3
Two Column Proof:
Statements Reasons
(1) (1) Given
(2) (2)
(3) (3)
(4) (4)
L is midpoint of KM
KL LM Definition of Midpoint
LM AB Given
KL AB of segments is transitive.
Given:
Prove:
MLK
BA
L is midpoint of KM
LM AB
KL AB
The given states that L is midpoint of KM and therefore KL LM by definition
of midpoint. We also have that the given states LM AB . This allow us to conclude
KL AB because of segments is transitive.
Convert to PARAGRAPH PROOF:
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9
STANDARDS 1,2,3
Given:
Prove:
MLK
BA
L is midpoint of KM
LM AB
KL AB
The given states that L is midpoint of KM and therefore KL LM by definition
of midpoint. We also have that the given states LM AB . This allow us to conclude
KL AB because of segments is transitive.
PARAGRAPH PROOF:
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10
B
CF
D
A
E
STANDARDS 1,2,3
Given:
EFD is right
Prove:
AFB CFBand are complementary.
Two Column Proof:
Statements Reasons
(1) (1)EFD is right Given
(2) (2)EC AD Definition of lines
(3) (3)AFC is right lines form 4 right s
(4) (4)AFC=m 90° Definition of right s
(5) (5)AFB +m AFCmCFB =m
(6) (6)
addition postulate
AFB +m CFB = 90°m Substitution prop. of (=)
AFB CFBand are complementary.
(7) (7) Definition of complementary s
B
CF
D
A
E
Review the following two column proof and then convert it to paragraph proof:
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11
STANDARDS 1,2,3
Given:
EFD is right
Prove:
AFB CFBand are complementary.
B
CF
D
A
E
Convert to PARAGRAPH PROOF:
From the given EFD is right
Two Column Proof:
Statements Reasons
(1) (1)EFD is right Given
(2) (2)EC AD Definition of lines
(3) (3)AFC is right
(4) (4)AFC=m 90° Definition of right s
(5) (5)AFB +m AFCmCFB =m
(6) (6)
addition postulate
AFB +m CFB = 90°m Substitution prop. of (=)
AFB CFBand are complementary.
(7) (7) Definition of complementary s
Lines form 4 right s
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12
STANDARDS 1,2,3
Given:
EFD is right
Prove:
AFB CFBand are complementary.
B
CF
D
A
E
Convert to PARAGRAPH PROOF:
From the given EFD is right and thus EC AD definition of linesby
Two Column Proof:
Statements Reasons
(1) (1)EFD is right Given
(2) (2)EC AD Definition of lines
(3) (3)AFC is right
(4) (4)AFC=m 90° Definition of right s
(5) (5)AFB +m AFCmCFB =m
(6) (6)
addition postulate
AFB +m CFB = 90°m Substitution prop. of (=)
AFB CFBand are complementary.
(7) (7) Definition of complementary s
Lines form 4 right s
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13
STANDARDS 1,2,3
Two Column Proof:
Statements Reasons
(1) (1)EFD is right Given
(2) (2)EC AD Definition of lines
(3) (3)AFC is right Lines form 4 right s
(4) (4)AFC=m 90° Definition of right s
(5) (5)AFB +m AFCmCFB =m
(6) (6)
addition postulate
AFB +m CFB = 90°m Substitution prop. of (=)
AFB CFBand are complementary.
(7) (7) Definition of complementary s
Given:
EFD is right
Prove:
AFB CFBand are complementary.
B
CF
D
A
E
Convert to PARAGRAPH PROOF:
From the given EFD is right and thus EC AD definition of lines.by
AFC is rightWe can also
observe that because lines form 4 right s
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14
STANDARDS 1,2,3
Two Column Proof:
Statements Reasons
(1) (1)EFD is right Given
(2) (2)EC AD Definition of lines
(3) (3)AFC is right Lines form 4 right s
(4) (4)AFC=m 90° Definition of right s
(5) (5)AFB +m AFCmCFB =m
(6) (6)
addition postulate
AFB +m CFB = 90°m Substitution prop. of (=)
AFB CFBand are complementary.
(7) (7) Definition of complementary s
Given:
EFD is right
Prove:
AFB CFBand are complementary.
B
CF
D
A
E
Convert to PARAGRAPH PROOF:
From the given EFD is right and thus EC AD definition of lines.by
AFC is rightWe can also
observe that because lines form 4 right s and AFC=m 90° by
definition of right s .
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15
STANDARDS 1,2,3
Two Column Proof:
Statements Reasons
(1) (1)EFD is right Given
(2) (2)EC AD Definition of lines
(3) (3)AFC is right Lines form 4 right s
(4) (4)AFC=m 90° Definition of right s
(5) (5)AFB +m AFCmCFB =m
(6) (6)
addition postulate
AFB +m CFB = 90°m Substitution prop. of (=)
AFB CFBand are complementary.
(7) (7) Definition of complementary s
Given:
EFD is right
Prove:
AFB CFBand are complementary.
B
CF
D
A
E
Convert to PARAGRAPH PROOF:
From the given EFD is right and thus EC AD definition of lines.by
AFC is rightWe can also
observe that because lines form 4 right s and AFC=m 90° by
definition of right s .
Now by addition postulate, it is that AFB +m AFCmCFB =m
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
STANDARDS 1,2,3
Two Column Proof:
Statements Reasons
(1) (1)EFD is right Given
(2) (2)EC AD Definition of lines
(3) (3)AFC is right Lines form 4 right s
(4) (4)AFC=m 90° Definition of right s
(5) (5)AFB +m AFCmCFB =m
(6) (6)
addition postulate
AFB +m CFB = 90°m Substitution prop. of (=)
AFB CFBand are complementary.
(7) (7) Definition of complementary s
Given:
EFD is right
Prove:
AFB CFBand are complementary.
B
CF
D
A
E
Convert to PARAGRAPH PROOF:
From the given EFD is right and thus EC AD definition of lines.by
AFC is rightWe can also
observe that because lines form 4 right s and AFC=m 90° by
definition of right s .
Now by addition postulate, it is that AFB +m AFC,mCFB =m where by
substitution prop. of (=) results: AFB +m CFB = 90°,m
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
STANDARDS 1,2,3
Two Column Proof:
Statements Reasons
(1) (1)EFD is right Given
(2) (2)EC AD Definition of lines
(3) (3)AFC is right Lines form 4 right s
(4) (4)AFC=m 90° Definition of right s
(5) (5)AFB +m AFCmCFB =m
(6) (6)
addition postulate
AFB +m CFB = 90°m Substitution prop. of (=)
AFB CFBand are complementary.
(7) (7) Definition of complementary s
Given:
EFD is right
Prove:
AFB CFBand are complementary.
B
CF
D
A
E
Convert to PARAGRAPH PROOF:
From the given EFD is right and thus EC AD definition of lines.by
AFC is rightWe can also
observe that because lines form 4 right s and AFC=m 90° by
definition of right s .
Now by addition postulate, it is that AFB +m AFC,mCFB =m where by
substitution prop. of (=) results: AFB +m CFB = 90°,m and this last statement
allow us to prove that AFB CFBand are complementary, by definition of complementary s.
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18
STANDARDS 1,2,3
Given:
EFD is right
Prove:
AFB CFBand are complementary.
B
CF
D
A
E
From the given EFD is right and thus EC AD definition of lines.by
AFC is rightWe can also
observe that because lines form 4 right s and AFC=m 90° by
definition of right s .
Now by addition postulate, it is that AFB +m AFC,mCFB =m where by
substitution prop. of (=) results: AFB +m CFB = 90°,m and this last statement
allow us to prove that AFB CFBand are complementary, by definition of complementary s.
PARAGRAPH PROOF:
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19
STANDARDS 1,2,3
Given:
Prove:
Two Column Proof:
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
(8) (8)
(9) (9)
B
ACE
D
F
GH
CE bisects BCA
FGH ECA
FGH +m BCD = 180°m2( )
CE bisects BCA Given
BCE ECA Definition of bisector
BCE=m ECAm Definition of s
FGH ECA Given
FGH=m ECAm
BCE=m FGHm
Definition of s
of s is transitive
BCE +m BCD = 180°mECA +m addition postulate
FGH +m BCD = 180°mFGH +m
FGH +m BCD = 180°m2( )
Substitution prop. of (=)
Adding like terms
B
ACE
D
Review the following two column proof and then convert it to paragraph proof:
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
STANDARDS
B
ACE
D
F
GH
Given:
Prove:
CE bisects BCA
FGH ECA
FGH +m BCD = 180°m2( )
FGH +m BCD = 180°m2( )
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
(8) (8)
(9) (9)
CE bisects BCA Given
BCE ECA Definition of bisector
BCE=m ECAm Definition of s
FGH ECA Given
FGH=m ECAm
BCE=m FGHm
Definition of s
of s is transitive
BCE +m BCD = 180°mECA +m addition postulate
FGH +m BCD = 180°mFGH +m Substitution prop. of (=)
Adding like terms
As stated in the given CE bisects BCA,
Convert to PARAGRAPH PROOF:
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STANDARDS
B
ACE
D
F
GH
Given:
Prove:
CE bisects BCA
FGH ECA
FGH +m BCD = 180°m2( )
FGH +m BCD = 180°m2( )
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
(8) (8)
(9) (9)
CE bisects BCA Given
BCE ECA Definition of bisector
BCE=m ECAm Definition of s
FGH ECA Given
FGH=m ECAm
BCE=m FGHm
Definition of s
of s is transitive
BCE +m BCD = 180°mECA +m addition postulate
FGH +m BCD = 180°mFGH +m Substitution prop. of (=)
Adding like terms
As stated in the given CE bisects BCA, and as consequence BCE ECA bydefinition of bisector,
Convert to PARAGRAPH PROOF:
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STANDARDS
B
ACE
D
F
GH
Given:
Prove:
CE bisects BCA
FGH ECA
FGH +m BCD = 180°m2( )
FGH +m BCD = 180°m2( )
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
(8) (8)
(9) (9)
CE bisects BCA Given
BCE ECA Definition of bisector
BCE=m ECAm Definition of s
FGH ECA Given
FGH=m ECAm
BCE=m FGHm
Definition of s
of s is transitive
BCE +m BCD = 180°mECA +m addition postulate
FGH +m BCD = 180°mFGH +m Substitution prop. of (=)
Adding like terms
As stated in the given CE bisects BCA, and as consequence BCE ECA bydefinition of bisector, so that BCE=m ECAm by definition of s.
Convert to PARAGRAPH PROOF:
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STANDARDS
B
ACE
D
F
GH
Given:
Prove:
CE bisects BCA
FGH ECA
FGH +m BCD = 180°m2( )
FGH +m BCD = 180°m2( )
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
(8) (8)
(9) (9)
CE bisects BCA Given
BCE ECA Definition of bisector
BCE=m ECAm Definition of s
FGH ECA Given
FGH=m ECAm
BCE=m FGHm
Definition of s
of s is transitive
BCE +m BCD = 180°mECA +m addition postulate
FGH +m BCD = 180°mFGH +m Substitution prop. of (=)
Adding like terms
As stated in the given CE bisects BCA, and as consequence BCE ECA bydefinition of bisector, so that BCE=m ECAm by definition of s. Now
also from the given, we have that FGH ECA,
Convert to PARAGRAPH PROOF:
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STANDARDS
B
ACE
D
F
GH
Given:
Prove:
CE bisects BCA
FGH ECA
FGH +m BCD = 180°m2( )
FGH +m BCD = 180°m2( )
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
(8) (8)
(9) (9)
CE bisects BCA Given
BCE ECA Definition of bisector
BCE=m ECAm Definition of s
FGH ECA Given
FGH=m ECAm
BCE=m FGHm
Definition of s
of s is transitive
BCE +m BCD = 180°mECA +m addition postulate
FGH +m BCD = 180°mFGH +m Substitution prop. of (=)
Adding like terms
As stated in the given CE bisects BCA, and as consequence BCE ECA bydefinition of bisector, so that BCE=m ECAm by definition of s. Now
also from the given, we have that FGH ECA, which implies FGH=m ECAm
by definition of s.
Convert to PARAGRAPH PROOF:
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STANDARDS
B
ACE
D
F
GH
Given:
Prove:
CE bisects BCA
FGH ECA
FGH +m BCD = 180°m2( )
FGH +m BCD = 180°m2( )
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
(8) (8)
(9) (9)
CE bisects BCA Given
BCE ECA Definition of bisector
BCE=m ECAm Definition of s
FGH ECA Given
FGH=m ECAm
BCE=m FGHm
Definition of s
of s is transitive
BCE +m BCD = 180°mECA +m addition postulate
FGH +m BCD = 180°mFGH +m Substitution prop. of (=)
Adding like terms
As stated in the given CE bisects BCA, and as consequence BCE ECA bydefinition of bisector, so that BCE=m ECAm by definition of s. Now
also from the given, we have that FGH ECA, which implies FGH=m ECAm
by definition of s. This last statement allow us then by of s is transitive,to state that BCE=m FGH.m
Convert to PARAGRAPH PROOF:
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STANDARDS
B
ACE
D
F
GH
Given:
Prove:
CE bisects BCA
FGH ECA
FGH +m BCD = 180°m2( )
FGH +m BCD = 180°m2( )
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
(8) (8)
(9) (9)
CE bisects BCA Given
BCE ECA Definition of bisector
BCE=m ECAm Definition of s
FGH ECA Given
FGH=m ECAm
BCE=m FGHm
Definition of s
of s is transitive
BCE +m BCD = 180°mECA +m addition postulate
FGH +m BCD = 180°mFGH +m Substitution prop. of (=)
Adding like terms
As stated in the given CE bisects BCA, and as consequence BCE ECA bydefinition of bisector, so that BCE=m ECAm by definition of s. Now
also from the given, we have that FGH ECA, which implies FGH=m ECAm
by definition of s. This last statement allow us then by of s is transitive,to state that BCE=m FGH.m Now from the figure by
BCE +m BCD = 180°,mECA +m
addition postulate, we have
Convert to PARAGRAPH PROOF:
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STANDARDS
B
ACE
D
F
GH
Given:
Prove:
CE bisects BCA
FGH ECA
FGH +m BCD = 180°m2( )
FGH +m BCD = 180°m2( )
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
(8) (8)
(9) (9)
CE bisects BCA Given
BCE ECA Definition of bisector
BCE=m ECAm Definition of s
FGH ECA Given
FGH=m ECAm
BCE=m FGHm
Definition of s
of s is transitive
BCE +m BCD = 180°mECA +m addition postulate
FGH +m BCD = 180°mFGH +m Substitution prop. of (=)
Adding like terms
As stated in the given CE bisects BCA, and as consequence BCE ECA bydefinition of bisector, so that BCE=m ECAm by definition of s. Now
also from the given, we have that FGH ECA, which implies FGH=m ECAm
by definition of s. This last statement allow us then by of s is transitive,to state that BCE=m FGH.m Now from the figure by
BCE +m BCD = 180°,mECA +m
addition postulate, we haveand this is FGH +m BCD = 180°mFGH +m
by substitution prop. of (=).
Convert to PARAGRAPH PROOF:
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
STANDARDS
B
ACE
D
F
GH
Given:
Prove:
CE bisects BCA
FGH ECA
FGH +m BCD = 180°m2( )
FGH +m BCD = 180°m2( )
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
(8) (8)
(9) (9)
CE bisects BCA Given
BCE ECA Definition of bisector
BCE=m ECAm Definition of s
FGH ECA Given
FGH=m ECAm
BCE=m FGHm
Definition of s
of s is transitive
BCE +m BCD = 180°mECA +m addition postulate
FGH +m BCD = 180°mFGH +m Substitution prop. of (=)
Adding like terms
As stated in the given CE bisects BCA, and as consequence BCE ECA bydefinition of bisector, so that BCE=m ECAm by definition of s. Now
also from the given, we have that FGH ECA, which implies FGH=m ECAm
by definition of s. This last statement allow us then by of s is transitive,to state that BCE=m FGH.m Now from the figure by
BCE +m BCD = 180°,mECA +m
addition postulate, we haveand this is FGH +m BCD = 180°mFGH +m
by substitution prop. of (=). Adding like terms results FGH +m BCD = 180°.m2( )
Convert to PARAGRAPH PROOF:
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29
STANDARDS 1,2,3
Given:
Prove:
B
ACE
D
F
GH
CE bisects BCA
FGH ECA
FGH +m BCD = 180°m2( )
As stated in the given and as consequence BCE ECA bydefinition of bisector, so that BCE=m ECAm by definition of s. Now
also from the given, we have that FGH ECA, which implies FGH=m ECAm
by definition of s. This last statement allow us then by of s is transitive,to state that BCE=m FGH.m Now from the figure by
BCE +m BCD = 180°,mECA +m
addition postulate, we haveand this is FGH +m BCD = 180°mFGH +m
by substitution prop. of (=). Adding like terms results FGH +m BCD = 180°.m2( )
PARAGRAPH PROOF:
CE bisects BCA,
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30
STANDARDS 1,2,3
Given:
FBD is right
Prove:
ABF CBDand are complementary.
Two Column Proof:
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
F
EB
C
A
D
F
EB
C
A
D
FBD is right Given
FBD=m 90° Definition of right s
FBD +m CBD = 180°mABF +m addition postulate
CBD = 180°mABF +m 90° +
ABF +m CBD = 90°m
Substitution prop. of (=)
Subtraction prop. of (=)
ABF CBDand are complementary. Definition of complementary s
Review the following two column proof and then convert it to paragraph proof:
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31
STANDARDS 1,2,3
F
EB
C
A
DCBD = 180°mABF +m 90° +
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
FBD is right Given
FBD=m 90° Definition of right s
FBD +m CBD = 180°mABF +m addition postulate
ABF +m CBD = 90°m
Substitution prop. of (=)
Subtraction prop. of (=)
ABF CBDand are complementary. Definition of complementary s
Given:FBD is right
Prove:ABF CBDand
are complementary.
Convert to PARAGRAPH PROOF:
The given is telling us that FBD is right,
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
32
STANDARDS 1,2,3
F
EB
C
A
DCBD = 180°mABF +m 90° +
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
FBD is right Given
FBD=m 90° Definition of right s
FBD +m CBD = 180°mABF +m addition postulate
ABF +m CBD = 90°m
Substitution prop. of (=)
Subtraction prop. of (=)
ABF CBDand are complementary. Definition of complementary s
Given:FBD is right
Prove:ABF CBDand
are complementary.
Convert to PARAGRAPH PROOF:
The given is telling us that FBD is right, then FBD=m 90° by definition of right s ,
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
33
STANDARDS 1,2,3
F
EB
C
A
DCBD = 180°mABF +m 90° +
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
FBD is right Given
FBD=m 90° Definition of right s
FBD +m CBD = 180°mABF +m addition postulate
ABF +m CBD = 90°m
Substitution prop. of (=)
Subtraction prop. of (=)
ABF CBDand are complementary. Definition of complementary s
Given:FBD is right
Prove:ABF CBDand
are complementary.
Convert to PARAGRAPH PROOF:
The given is telling us that FBD is right, then FBD=m 90° by definition of right s ,and from the figure: FBD +m CBD = 180°mABF +m by addition postulate.
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
34
STANDARDS 1,2,3
F
EB
C
A
DCBD = 180°mABF +m 90° +
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
FBD is right Given
FBD=m 90° Definition of right s
FBD +m CBD = 180°mABF +m addition postulate
ABF +m CBD = 90°m
Substitution prop. of (=)
Subtraction prop. of (=)
ABF CBDand are complementary. Definition of complementary s
Given:FBD is right
Prove:ABF CBDand
are complementary.
Convert to PARAGRAPH PROOF:
The given is telling us that FBD is right, then FBD=m 90° by definition of right s ,and from the figure: FBD +m CBD = 180°mABF +m by addition postulate.
We can then have CBD = 180°mABF +m 90° + using substitution prop. of (=).
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
35
STANDARDS 1,2,3
F
EB
C
A
DCBD = 180°mABF +m 90° +
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
FBD is right Given
FBD=m 90° Definition of right s
FBD +m CBD = 180°mABF +m addition postulate
ABF +m CBD = 90°m
Substitution prop. of (=)
Subtraction prop. of (=)
ABF CBDand are complementary. Definition of complementary s
Given:FBD is right
Prove:ABF CBDand
are complementary.
Convert to PARAGRAPH PROOF:
The given is telling us that FBD is right, then FBD=m 90° by definition of right s ,and from the figure: FBD +m CBD = 180°mABF +m by addition postulate.
We can then have CBD = 180°mABF +m 90° + using substitution prop. of (=).
Now, if we subtract 90° from both sides applying subtraction prop. of (=), this results inABF +m CBD = 90°m which is what we intended to prove, and that enable us to state
that
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
36
STANDARDS 1,2,3
F
EB
C
A
DCBD = 180°mABF +m 90° +
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
FBD is right Given
FBD=m 90° Definition of right s
FBD +m CBD = 180°mABF +m addition postulate
ABF +m CBD = 90°m
Substitution prop. of (=)
Subtraction prop. of (=)
ABF CBDand are complementary. Definition of complementary s
Given:FBD is right
Prove:ABF CBDand
are complementary.
Convert to PARAGRAPH PROOF:
The given is telling us that FBD is right, then FBD=m 90° by definition of right s ,and from the figure: FBD +m CBD = 180°mABF +m by addition postulate.
We can then have CBD = 180°mABF +m 90° + using substitution prop. of (=).
Now, if we subtract 90° from both sides applying subtraction prop. of (=), this results inABF +m CBD = 90°m which is what we intended to prove, and that enable us to state
that ABF CBDand are complementary, by definition of complementary s.
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
37
STANDARDS 1,2,3
Given:
FBD is right
Prove:
ABF CBDand are complementary.
F
EB
C
A
D
The given is telling us that FBD is right, then FBD=m 90° by definition of right s ,and from the figure: FBD +m CBD = 180°mABF +m by addition postulate.
We can then have CBD = 180°mABF +m 90° + using substitution prop. of (=).
Now, if we subtract 90° from both sides applying subtraction prop. of (=), this results inABF +m CBD = 90°m which is what we intended to prove, and that enable us to state
that ABF CBDand are complementary, by definition of complementary s.
PARAGRAPH PROOF:
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
38STANDARDS 1,2,3
Given:
Prove:
Two Column Proof:
Statements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
CAB
H
G
D E F
CAB
H
G
D E FGE is a transversalAC and DF are
GBC FEHand are supplementary.
GE is a transversal
AC and DF are Given
GBC CBEand are a linear pair Definition of linear pair
GBC +m CBE = 180°m s in a linear pair are supplementary
CBE FEH In lines cut by a transversal CORRESPONDING s are
Definition of sCBE=m FEHm
GBC +m FEH = 180°m Substitution prop. of (=)
GBC FEHand are supplementary.
Definition of supplementary s
Review the following two column proof and then convert it to paragraph proof:
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
39
STANDARDS 1,2,3
CAB
H
G
D E FStatements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
GE is a transversal
AC and DF are Given
GBC CBEand are a linear pair Definition of linear pair
GBC +m CBE = 180°m s in a linear pair are supplementary
CBE FEH In lines cut by a transversal CORRESPONDING s are
Definition of sCBE=m FEHm
GBC +m FEH = 180°m Substitution prop. of (=)
GBC FEHand are supplementary.
Definition of supplementary s
Given:
Prove:GE is a transversalAC and DF are
GBC FEHandare supplementary.
The given states that AC and DF are and GE is a transversal,
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
40
STANDARDS 1,2,3
CAB
H
G
D E FStatements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
GE is a transversal
AC and DF are Given
GBC CBEand are a linear pair Definition of linear pair
GBC +m CBE = 180°m s in a linear pair are supplementary
CBE FEH In lines cut by a transversal CORRESPONDING s are
Definition of sCBE=m FEHm
GBC +m FEH = 180°m Substitution prop. of (=)
GBC FEHand are supplementary.
Definition of supplementary s
Given:
Prove:GE is a transversalAC and DF are
GBC FEHandare supplementary.
The given states that AC and DF are and GE is a transversal, therefore we have thatCBE FEH because in lines cut by a transversal CORRESPONDING s are ;
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
41
STANDARDS 1,2,3
CAB
H
G
D E FStatements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
GE is a transversal
AC and DF are Given
GBC CBEand are a linear pair Definition of linear pair
GBC +m CBE = 180°m s in a linear pair are supplementary
CBE FEH In lines cut by a transversal CORRESPONDING s are
Definition of sCBE=m FEHm
GBC +m FEH = 180°m Substitution prop. of (=)
GBC FEHand are supplementary.
Definition of supplementary s
Given:
Prove:GE is a transversalAC and DF are
GBC FEHandare supplementary.
The given states that AC and DF are and GE is a transversal, therefore we have thatCBE FEH because in lines cut by a transversal CORRESPONDING s are ;
and this implies CBE=m FEHm by definition of s.
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
42
STANDARDS 1,2,3
CAB
H
G
D E FStatements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
GE is a transversal
AC and DF are Given
GBC CBEand are a linear pair Definition of linear pair
GBC +m CBE = 180°m s in a linear pair are supplementary
CBE FEH In lines cut by a transversal CORRESPONDING s are
Definition of sCBE=m FEHm
GBC +m FEH = 180°m Substitution prop. of (=)
GBC FEHand are supplementary.
Definition of supplementary s
Given:
Prove:GE is a transversalAC and DF are
GBC FEHandare supplementary.
The given states that AC and DF are and GE is a transversal, therefore we have thatCBE FEH because in lines cut by a transversal CORRESPONDING s are ;
Now, we also have that
GBC CBEand are a linear pair by def. of linear pair;
and this implies CBE=m FEHm by definition of s.
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
43
STANDARDS 1,2,3
CAB
H
G
D E FStatements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
GE is a transversal
AC and DF are Given
GBC CBEand are a linear pair Definition of linear pair
GBC +m CBE = 180°m s in a linear pair are supplementary
CBE FEH In lines cut by a transversal CORRESPONDING s are
Definition of sCBE=m FEHm
GBC +m FEH = 180°m Substitution prop. of (=)
GBC FEHand are supplementary.
Definition of supplementary s
Given:
Prove:GE is a transversalAC and DF are
GBC FEHandare supplementary.
The given states that AC and DF are and GE is a transversal, therefore we have thatCBE FEH because in lines cut by a transversal CORRESPONDING s are ;
Now, we also have that
GBC CBEand are a linear pair by def. of linear pair;
and this implies CBE=m FEHm by definition of s.
so GBC +m CBE = 180°m
because s in a linear pair are supplementary;
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
44
STANDARDS 1,2,3
CAB
H
G
D E FStatements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
GE is a transversal
AC and DF are Given
GBC CBEand are a linear pair Definition of linear pair
GBC +m CBE = 180°m s in a linear pair are supplementary
CBE FEH In lines cut by a transversal CORRESPONDING s are
Definition of sCBE=m FEHm
GBC +m FEH = 180°m Substitution prop. of (=)
GBC FEHand are supplementary.
Definition of supplementary s
Given:
Prove:GE is a transversalAC and DF are
GBC FEHandare supplementary.
The given states that AC and DF are and GE is a transversal, therefore we have thatCBE FEH because in lines cut by a transversal CORRESPONDING s are ;
Now, we also have that
GBC CBEand are a linear pair by def. of linear pair;
and this implies CBE=m FEHm by definition of s.
so GBC +m CBE = 180°m
because s in a linear pair are supplementary;this becomes GBC +m FEH = 180°m
by substitution prop. of (=).
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
45
STANDARDS 1,2,3
CAB
H
G
D E FStatements Reasons
(1) (1)
(2) (2)
(3) (3)
(4) (4)
(5) (5)
(6) (6)
(7) (7)
GE is a transversal
AC and DF are Given
GBC CBEand are a linear pair Definition of linear pair
GBC +m CBE = 180°m s in a linear pair are supplementary
CBE FEH In lines cut by a transversal CORRESPONDING s are
Definition of sCBE=m FEHm
GBC +m FEH = 180°m Substitution prop. of (=)
GBC FEHand are supplementary.
Definition of supplementary s
Given:
Prove:GE is a transversalAC and DF are
GBC FEHandare supplementary.
The given states that AC and DF are and GE is a transversal, therefore we have thatCBE FEH because in lines cut by a transversal CORRESPONDING s are ;
Now, we also have that
GBC CBEand are a linear pair by def. of linear pair;
and this implies CBE=m FEHm by definition of s.
so GBC +m CBE = 180°m
because s in a linear pair are supplementary;this becomes GBC +m FEH = 180°m
by substitution prop. of (=). We can conclude GBC FEHand are supplementary,
since they comply with definition of supplementary s. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
46STANDARDS 1,2,3
Given:
Prove:
CAB
H
G
D E FGE is a transversalAC and DF are
GBC FEHand are supplementary.
The given states that AC and DF are and GE is a transversal, therefore we have thatCBE FEH because in lines cut by a transversal CORRESPONDING s are ;
Now, we also have that
GBC CBEand are a linear pair by def. of linear pair;
and this implies CBE=m FEHm by definition of s.
so GBC +m CBE = 180°m
because s in a linear pair are supplementary;this becomes GBC +m FEH = 180°m
by substitution prop. of (=). We can conclude GBC FEHand are supplementary,
since they comply with definition of supplementary s.
PARAGRAPH PROOF:
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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