1 3 Parr a Graph Proofs Intro

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

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

http://www.mrperezonlinemathtutor.com/

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.



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.



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



5

STANDARDS 1,2,3

Two Column Proof:

Statements Reasons

(1) (1) Given

(2) (2)

(3) (3)

(4) (4)

L is midpoint of KM


LM AB Given


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:



6

STANDARDS 1,2,3

Two Column Proof:

Statements Reasons

(1) (1) Given

(2) (2)

(3) (3)

(4) (4)

L is midpoint of KM


LM AB Given


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.




7

STANDARDS 1,2,3

Two Column Proof:

Statements Reasons

(1) (1) Given

(2) (2)

(3) (3)

(4) (4)

L is midpoint of KM


LM AB Given


Given:

Prove:

MLK

BA

L is midpoint of KM

LM AB

KL AB


of midpoint. We also have that the given states LM AB




8

STANDARDS 1,2,3

Two Column Proof:

Statements Reasons

(1) (1) Given

(2) (2)

(3) (3)

(4) (4)

L is midpoint of KM


LM AB Given


Given:

Prove:

MLK

BA

L is midpoint of KM

LM AB

KL AB


of midpoint. We also have that the given states LM AB . This allow us to conclude

KL AB because of segments is transitive.




9

STANDARDS 1,2,3

Given:

Prove:

MLK

BA

L is midpoint of KM

LM AB

KL AB


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:



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:



11

STANDARDS 1,2,3

Given:

EFD is right

Prove:

AFB CFBand are

complementary.

B

CF

D

A

E


From the given EFD is right

Two Column Proof:

Statements Reasons



(3) (3)AFC is right



(6) (6)

addition postulate


AFB CFBand are

complementary.


complementary s

Lines form 4 right s



12

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 linesby

Two Column Proof:

Statements Reasons



(3) (3)AFC is right



(6) (6)

addition postulate


AFB CFBand are

complementary.


complementary s

Lines form 4 right s



13

STANDARDS 1,2,3

Two Column Proof:

Statements Reasons



(3) (3)AFC is right Lines form 4 right s



(6) (6)

addition postulate


AFB CFBand are

complementary.


complementary s

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 right

We can also

observe that because lines form 4 right s



14

STANDARDS 1,2,3

Two Column Proof:

Statements Reasons






(6) (6)

addition postulate


AFB CFBand are

complementary.


complementary s

Given:

EFD is right

Prove:

AFB CFBand are

complementary.

B

CF

D

A

E



AFC is right

We can also

observe that because lines form 4 right s and AFC=m 90° by

definition of right s .



15

STANDARDS 1,2,3

Two Column Proof:

Statements Reasons






(6) (6)

addition postulate


AFB CFBand are

complementary.


complementary s

Given:

EFD is right

Prove:

AFB CFBand are

complementary.

B

CF

D

A

E



AFC is right

We can also



Now by addition postulate, it is that AFB +m AFCmCFB =m



STANDARDS 1,2,3

Two Column Proof:

Statements Reasons






(6) (6)

addition postulate


AFB CFBand are

complementary.


complementary s

Given:

EFD is right

Prove:

AFB CFBand are

complementary.

B

CF

D

A

E



AFC is right

We can also



Now by addition postulate, it is that AFB +m AFC,mCFB =m where by

substitution prop. of (=) results: AFB +m CFB = 90°,m



STANDARDS 1,2,3

Two Column Proof:

Statements Reasons






(6) (6)

addition postulate


AFB CFBand are

complementary.


complementary s

Given:

EFD is right

Prove:

AFB CFBand are

complementary.

B

CF

D

A

E



AFC is right

We can also




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.PRESENTATION CREATED BY SIMON PEREZ. All rights reserved


18

STANDARDS 1,2,3

Given:

EFD is right

Prove:

AFB CFBand are complementary.

B

CF

D

A

E


AFC is right

We can also




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:



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:



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)




FGH ECA Given

FGH=m ECAm

BCE=m FGHm

Definition of s

of s is transitive


FGH +m BCD = 180°mFGH +m Substitution prop. of (=)

Adding like terms

As stated in the given CE bisects BCA,




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)




FGH ECA Given

FGH=m ECAm

BCE=m FGHm

Definition of s

of s is transitive



Adding like terms

As stated in the given CE bisects BCA, and as consequence BCE ECA by

definition of bisector,




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)




FGH ECA Given

FGH=m ECAm

BCE=m FGHm

Definition of s

of s is transitive



Adding like terms


definition of bisector, so that BCE=m ECAm by definition of s.




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)




FGH ECA Given

FGH=m ECAm

BCE=m FGHm

Definition of s

of s is transitive



Adding like terms


definition of bisector, so that BCE=m ECAm by definition of s. Now

also from the given, we have that FGH ECA,




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)




FGH ECA Given

FGH=m ECAm

BCE=m FGHm

Definition of s

of s is transitive



Adding like terms



also from the given, we have that FGH ECA, which implies FGH=m ECAm

bydefinition of s.




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)




FGH ECA Given

FGH=m ECAm

BCE=m FGHm

Definition of s

of s is transitive



Adding like terms




bydefinition of s. This last statement allow us then by of s is transitive,

to state that BCE=m FGH.m




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)




FGH ECA Given

FGH=m ECAm

BCE=m FGHm

Definition of s

of s is transitive



Adding like terms





to state that BCE=m FGH.m Now from the figure by

BCE +m BCD = 180°,mECA +m

addition postulate, we have




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)




FGH ECA Given

FGH=m ECAm

BCE=m FGHm

Definition of s

of s is transitive



Adding like terms








and this is FGH +m BCD = 180°mFGH +m

by substitution prop. of (=).




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)




FGH ECA Given

FGH=m ECAm

BCE=m FGHm

Definition of s

of s is transitive



Adding like terms









by substitution prop. of (=). Adding like terms results FGH +m BCD = 180°.m2( )




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 by








by substitution prop. of (=). Adding like terms results FGH +m BCD = 180°.m2( )

PARAGRAPH PROOF:

CE bisects BCA,



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


Subtraction prop. of (=)

ABF CBDand are complementary. Definition of

complementary s

Review the following two column proof and then convert it

to paragraph proof:



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



ABF +m CBD = 90°m




complementary s

Given:

FBD is rightProve:

ABF CBDand

are complementary.


The given is telling us that FBD is right,



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



ABF +m CBD = 90°m




complementary s

Given:

FBD is rightProve:

ABF CBDand

are complementary.


The given is telling us that FBD is right, then FBD=m 90° by definition of right s ,



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



ABF +m CBD = 90°m




complementary s

Given:

FBD is rightProve:

ABF CBDand

are complementary.



and from the figure: FBD +m CBD = 180°mABF +m by addition postulate.



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



ABF +m CBD = 90°m




complementary s

Given:

FBD is rightProve:

ABF CBDand

are complementary.




We can then have CBD = 180°mABF +m 90° + using substitution prop. of (=).



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



ABF +m CBD = 90°m




complementary s

Given:

FBD is rightProve:

ABF CBDand

are complementary.





Now, if we subtract 90° from both sides applying subtraction prop. of (=), this results in

ABF +m CBD = 90°m which is what we intended to prove, and that enable us to state

that



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



ABF +m CBD = 90°m




complementary s

Given:

FBD is rightProve:

ABF CBDand

are complementary.







that ABF CBDand are complementary, by definition of complementary s.



37

STANDARDS 1,2,3

Given:

FBD is right

Prove:

ABF CBDand are complementary.

F

EB

C

A

D






that ABF CBDand are complementary, by definition of complementary s.

PARAGRAPH PROOF:



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)

CA

B

H

G

D E F

CA

B

H

G

D E FGE is a transversal

AC 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:



39

STANDARDS 1,2,3

CA

B

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




CORRESPONDING s are



GBC FEHand are

supplementary.

Definition of

supplementary s

Given:

Prove:

GE is a transversal

AC and DF are

GBC FEHand

are supplementary.

The given states that AC and DF are and GE is a transversal,



40

STANDARDS 1,2,3

CA

B

H

G


(1) (1)

(2) (2)

(3) (3)

(4) (4)

(5) (5)

(6) (6)

(7) (7)

GE is a transversal

AC and DF are Given




CORRESPONDING s are



GBC FEHand are

supplementary.

Definition of

supplementary s

Given:

Prove:

GE is a transversal

AC and DF are

GBC FEHand

are supplementary.

The given states that AC and DF are and GE is a transversal, therefore we have that

CBE FEH because in lines cut by a transversal CORRESPONDING s are ;



41

STANDARDS 1,2,3

CA

B

H

G


(1) (1)

(2) (2)

(3) (3)

(4) (4)

(5) (5)

(6) (6)

(7) (7)

GE is a transversal

AC and DF are Given




CORRESPONDING s are



GBC FEHand are

supplementary.

Definition of

supplementary s

Given:

Prove:

GE is a transversal

AC and DF are

GBC FEHand

are supplementary.



and this implies CBE=m FEHm by definition of s.



42

STANDARDS 1,2,3

CA

B

H

G


(1) (1)

(2) (2)

(3) (3)

(4) (4)

(5) (5)

(6) (6)

(7) (7)

GE is a transversal

AC and DF are Given




CORRESPONDING s are



GBC FEHand are

supplementary.

Definition of

supplementary s

Given:

Prove:

GE is a transversal

AC and DF are

GBC FEHand

are supplementary.



Now, we also have that

GBC CBEand are a linear pair by def. of linear pair;




43

STANDARDS 1,2,3

CA

B

H

G


(1) (1)

(2) (2)

(3) (3)

(4) (4)

(5) (5)

(6) (6)

(7) (7)

GE is a transversal

AC and DF are Given




CORRESPONDING s are



GBC FEHand are

supplementary.

Definition of

supplementary s

Given:

Prove:

GE is a transversal

AC and DF are

GBC FEHand

are supplementary.






so GBC +m CBE = 180°m

because s in a linear pair are supplementary;



44

STANDARDS 1,2,3

CA

B

H

G


(1) (1)

(2) (2)

(3) (3)

(4) (4)

(5) (5)

(6) (6)

(7) (7)

GE is a transversal

AC and DF are Given




CORRESPONDING s are



GBC FEHand are

supplementary.

Definition of

supplementary s

Given:

Prove:

GE is a transversal

AC and DF are

GBC FEHand

are supplementary.







because s in a linear pair are supplementary; this becomes GBC +m FEH = 180°m

by substitution prop. of (=).



45

STANDARDS 1,2,3

CA

B

H

G


(1) (1)

(2) (2)

(3) (3)

(4) (4)

(5) (5)

(6) (6)

(7) (7)

GE is a transversal

AC and DF are Given




CORRESPONDING s are



GBC FEHand are

supplementary.

Definition of

supplementary s

Given:

Prove:

GE is a transversal

AC and DF are

GBC FEHand

are supplementary.








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:

CA

B

H

G

D E FGE is a transversal

AC and DF are

GBC FEHand are supplementary.








by substitution prop. of (=). We can conclude GBC FEHand are supplementary,

since they comply with definition of supplementary s.

PARAGRAPH PROOF:



Documents

1 3 Parr a Graph Proofs Intro