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SSS/SAS/ASA Proofs. C. !. 1. 2. A. B. D. Quiz Question. W. !. 3. 4. Y. Z. X. !. G. I. H. J. F. K. L. C. !. A. 1. D. 2. B. M. !. R. N. O. !. B. E. C. 1. 2. A. D. Quiz Question. !. N. Q. 4. 5. O. P. M. C. 1. 2. B. A. D. E. G. H. 1. 2. F.

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slide2

C

!

1

2

A

B

D

slide4

!

G

I

H

J

F

K

L

slide5

C

!

A

1

D

2

B

slide6

M

!

R

N

O

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!

B

E

C

1

2

A

D

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C

1

2

B

A

D

E

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G

H

1

2

F

E

C

B

D

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B

E

C

1

2

A

D

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B

D

F

1

2

A

C

3

4

E

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A

X

Y

1

3

B

2

4

C

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A

X

Y

1

3

2

4

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S

3

1

X

R

T

2

4

U

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C

D

1

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4

A

B

Y

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E

F

G

C

D

A

B

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S

R

Q

P

T

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B

A

C

Hw 11.3

D

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C

E

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Hw 11.3

D

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F

G

D

E

A

B

C

Hw 11.4

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U

T

S

W

V

R

Hw 11.4

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N

R

Hw 11.5

M

S

P

Q

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D

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C

Hw 11.5

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A

Hw 11.6

1

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M

C

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QUIZ

Z

Y

2

3

4

1

W

X

triangle congruence and similarity proofs

Triangle Congruence and Similarity Proofs

Augustus Vitug

Melissa Moize

proof title
Sketch

AB  AC

BD  DC

AD  AD

BAD  CAD

Proof Title:

Statement

Reason

  • Given
  • Given
  • Reflexive property
  • SSS

Given

AB  AC

BD  DC

Prove

BAD  CAD

proof title1
Sketch

ABC DCB

CB  BC

DBC  ACB

ABC  DCB

CA  DB

AB  CD

Proof Title:

Statement

Reason

  • Alt. Int. ’s
  • reflexive
  • Alt. Int. ’s
  • ASA
  • CPCTC
  • CPCTC

Given

ABCD

Prove

AC  BD

CD  AB

proof title parallelogram diagonals bisect each other
Sketch

ABC  DCB

CB  BC

DBC  ACB

 ABC   DCB

AB  CD

ADC  BAD

 ABE  DCF

AE DE

BECE

Proof Title: Parallelogram Diagonals Bisect Each Other

Statement

Reason

  • Alt. Int. ’s
  • Reflexive property
  • Alt. Int. ’s
  • ASA
  • CPCTC
  • Alt. Int. ’s
  • ASA
  • CPCTC
  • CPCTC

Given

ABCD

Prove

AEDE

BECE

proof title2
Sketch

ABC  DCB

CB  BC

ACB  DBC

ABC  DCB

A  D

ABC + DBC  DCB +ACB

Proof Title:

Statement

Reason

  • Alt. Int. ’s
  • reflexive
  • Alt. Int. ’s
  • ASA
  • CPCTC
  • 2nd Postulate

Given

Prove

A  D

ABD  DCA

proof title compound polygon 1
Sketch

ABC DCB

CBBC

CBD BCA

ABC  DCB

FBD+BFD+BDF  ECA+CEA+CAE

BDF CAE

AEFD

ACBD

BFD  CEA

CE BF

EF  FE

EF+FB  FE+EC

EB  CF

AB  CD

AEFD

CFD  BEA

Proof Title: Compound Polygon #1

Statement

Reason

  • Alt. int. ’s
  • reflexive property
  • Alt. int. ’s
  • ASA
  • Triangle Sums
  • Subtraction Property
  • Given
  • CPCTC
  • ASA
  • CPCTC
  • Reflexive Property
  • Sums of equals
  • Sums of equals
  • CPCTC
  • Given
  • SSS

Given

ABCD

AEFD

AEC=90

BFD=90

Prove

CFD  BEA

proof title compound 2
Sketch

AC BD

ACD BDC

CD  DC

ADC  BDC

CE+BE  DE+AE

CE  DE

CED is isos

Proof Title: Compound # 2

Statement

Reason

  • Given
  • Def. isos
  • Reflexive
  • SAS
  • Sums of Equals
  • Subtraction property
  • def. isos.

Given

ABCD is iso.

Prove

CED is isos

proof title3
Sketch

AB  CD

B  D

BE  DE

BEA  DEC

G   H

EC  EA

FE  XE

FEA  XEC

AF  CX

Proof Title: ?

Statement

Reason

  • Given
  • Given
  • Def. of Midpoint
  • SAS
  • CPCTC
  • CPCTC
  • Def. of Midpoint
  • SAS
  • CPCTC

Given

AB  CD

B  D

E is the midpoint of BD & FX

Prove

AF  CX

proof title simple bowtie
Sketch

AECE

DEC  BEA

BE DE

AEB  CED

ABE  CDE

ABE,CDE are alt. int. ’s

AB || DC

Proof Title: Simple Bowtie

Statement

Reason

  • Def. Bisector
  • Vertical ’s
  • Def. Bisector .
  • SAS
  • CPCTC
  • Def. Alt. Int. ’s
  •  Alt. Int. ’s are congruent

Given

Line AC bis. Line BD

Prove

AB || CD

proof title4
Sketch

A  D

BC  FE

B = 90- A  E = 90- D

B  E

ABC  DEF

BA  DE

Proof Title:

Statement

Reason

  • given
  • given
  • difference of equals
  • - property
  • ASA
  • CPCTC

Given

ABC is a Right 

BC  FE

A  D

Prove

BA  DE

proof title compound bowtie 1
Sketch

AGE+EGB CGF+ FGD

AGB  CGD

FGD  EGB

ABGCDG

DFG  BEG

Proof Title: Compound Bowtie #1

Statement

Reason

  • Given
  • Vertical ’s
  • Given
  • SAS
  • Vertical ’s
  • Given
  • Alt. Int. ’s
  • ASA
  • CPCTC

Given

Prove

proof title5
Sketch

AED  BEC

DE  CE

ADC-EDC  BCD ECD

ADE  BCE

ADE  BCE

AD  BC

ADC   BCD

DC  CD

ADC  BCD

Proof Title:

Statement

Reason

  • Vertical ’s
  • Def. iso. 
  • Dif. Of equals
  • Sub. Prop.
  • ASA
  • CPCTC
  • Sum of Equals
  • Reflexive Prop.
  • SAS

Given

EDC is isosceles

Prove

ADC  BCD

proof title6
Sketch

AE  BE

AED  BEC

DE  CE

ADE  BCE

FDE  GCE

DECE

FED  GEC

FD  GC

EDC  ECD

ADE+EDC  BCE+ ECD

ADCBCD

FGDC

Proof Title:

Statement

Reason

  • Given
  • Vert. ’s
  • def. iso. 
  • SAS
  • CPCTC
  • def. iso. 
  • alt. int. ’s
  • CPCTC
  • Def. iso. 
  • Addition prop. Of equality

Given

AEBE

EDC is isosceles

Prove

FG DC

proof title diagonals of a rectangle
Sketch

AC  BD

ACD  BDC

CD  DC

 ACD   BDC

AD  BC

Proof Title: Diagonals of a Rectangle

Statement

Reason

  • Dfn. Rectangle
  • Dfn. Rectangle
  • Reflexive
  • SAS
  • CPCTC

Given

Rectangle ABCD

Prove

AD  BC

proof title7
Sketch

BAM  CAM

ABAC

B C

BAM  CAM

BM  CM

BMA  CMA

BMC  180

BMA = 1/2180=90= CMA

AM  bis. Of BC

Proof Title:

Statement

Reason

  •  bis.
  • Given
  • Def. iso. 
  • ASA
  • CPCTC
  • CPCTC
  • straight ’s
  • Angle Addition

Given

ABC w/

AM  bis.

ABAC

Prove

AM bis.

proof title8
Sketch

¾=5/6.6666

AD/AB=AE/AC

A  A

ADE  ABC

Proof Title:

Statement

Reason

  • 36&2/3=4  5= 20
  • CPSTP
  • Reflexive
  • SAS

Given

DE || BC

Prove

ADE  ABC

proof title simple butterfly
Sketch

AEBE

BEDAEC

CE  DE

AC  BD

Proof Title: Simple butterfly

Statement

Reason

  • Given
  • Vertical angels
  • Given
  • CPCTC

Given

AEBE

CEDE

Prove

AC  BD

proof title9
Sketch

BCA  EFD

FC  CF

AF  CD

AF + FC  DC + CF

ABC  DEF

Proof Title:

Statement

Reason

  • Alt. Int. ’s
  • Reflexive
  • Given
  • Addition
  • ASA

Given

AF  DC

FE || CB

AB || DE

Prove

ABC  DEF

proof title10
Sketch

BG  EG

EG + GF=BG+GC

BC  EF

FC  FC

AF+FC=DC+CF

GFC  GCF

AC=CD

ABC  DEF

Proof Title:

Statement

Reason

  • Given
  • Sum’s of =‘s
  • Segment addition
  • reflexive
  • Sum’s of =‘s
  • def. isosceles 
  • Segment Addition
  • SAS

Given

FGC Isosceles

AF  DC

BG  EG

Prove

ABC  DEF

proof title isos triangle to bisector
Sketch

BM  CM

AMB = 90 = AMC

AB  AC

B  C

BAM  CAM

BAM   CAM

AM is  bis. of A

Proof Title: Isos. Triangle  to  bisector

Statement

Reason

  • Def.  bis.
  • Def.  bis.
  • given
  • def. isos. triangle
  • SAS
  • CPCTC
  • Given
  • ABC is isos.

AB  AC

AM is 

Prove

 bis. of BC is  bis. Of A

proof title11
Sketch

BMA=90= CMA

B  C

B+ BMA+ BAM=180

C+ CMA+ CAM =180

B+ BMA+ BAM= C+ CMA+ CAM

BAM  CAM

 AM is bis. of A

Proof Title:

Statement

Reason

  • Def. Alt.
  • isos triangle
  •  sums
  •  sums
  • transitive
  • - property

Given

ABC is isos.

AC  AB

AM is Alt.

Prove

Alt. AM is  bis. of A

proof title12
Sketch

AB  AC

C  B

AMB=90= AMC

BAM  CAM

BM  CM

Alt. AM is  bis. of BC

Proof Title:

Statement

Reason

  • given
  • def. of isos.
  • def. Alt.
  •  sums
  • ASA
  • CPCTC

Given

ABC is isos.

AB  AC

AM is Alt.

Prove

Alt. AM is  bis. of BC

proof title bisectors of an isosceles trapezoid
Sketch

ACD   BDC

CD  DC

CEA  DEB

AC  BD

C D

CE  DE

ACE BED

AE  BE

Proof Title: Bisectors Of An Isosceles Trapezoid

Statement

Reason

  • Base ’s
  • Reflexive Property
  • Alternate Interior ’s
  • Definition Isosceles 
  • Base ’s
  • Definition of Midpoint
  • SAS

Given

Trapezoid ABCD is and isosceles 

E is midpoint of CD

Prove

AE  BE

proof title pythagorean theorem
Sketch

ABC is a Rt. Triangle

CD is the altitude to hyp.

c/a=a/e

a2=c e

c/b=b/d

b2=cd

a2 + b2 = ce + cd

a2+b2=c2

Proof Title: Pythagorean Theorem

Statement

Reason

  • Given
  • Through a pt. there is 1 line  to a given line
  • Either leg is geom. Mean bet. Hyp. And adj. seg.
  • Multi proportion prop.
  • Either leg is geom. Mean bet. Hyp and adj. seg.
  • Multi. Propportion prop.
  • Distributive Prop.
  • Substitution

Given

Rt. ABC

Prove

a2+b2=c2

proof title13
Sketch

 DAB   CAB

 DBA   CBA

AB  AB

 ABD  ABC

Proof Title:

Statement

Reason

  • Given
  • Given
  • Reflexive Property
  • ASA

Given

DAB  CAB

DBA  CBA

Prove

 ABD   ABC

proof title compound butterfly 1
Sketch

AE  FE

AED  FEJ

DE  JE

ADE  FJE

D  J

AEB+BEC+CED  FEG+GEH+HEJ

CED  HEJ

CED  HEJ

BEC  GEH

CE  HE

BCE +DCE=180

GHE +JHE=180

BCE+DCE  GHE+ JHE

BCE  GHE

BEC  GHE

Proof Title: Compound Butterfly #1

Statement

Reason

  • given
  • vert. ’s
  • given
  • SAS
  • CPCTC
  • sums ’s
  • - prop
  • ASA
  • given
  • CPCTC
  • supp. ’s
  • supp. ’s
  • sums ’s
  • - prop
  • ASA

Given

AE  FE, DE  JE

BEC  GEH

AEB  FEG

Prove

BEC  GEH

proof title14
SketchProof Title:

Statement

Reason

Given

Prove

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