Transcript chapter four proofs

ASA- Angle Side Angle
Used to prove triangle congruence: if two angles and
the included side of two triangles are congruent,
then the triangles are congruent.
K
L
Given: KL and NO are parallel; M bisects KO.
Prove:
M
KLM ≡
Statements
Reasons
KL and NO are parallel; M
bisects KO.
KML ≡
OMN
∟MKL ≡ ∟MON
KM ≡ MO
Given
KLM ≡
N
O
ONM
ONM
Vertical Angles
Alt. Interior
Definition of bisect
ASA
Angle Angle Side -Two triangles can be proven to
be congruent if two angles and the not included
side are congruent.
Statements
Reasons
DE=FG; DA ll EC;
<B and <E are
right angles
Given
<A = 90°
<C = 90°
Definition right
angle
<A = <C
Transitive
Property
EF = EF
Reflexive Property
DF = GE
Overlapping
Segments
<D = <E
Corresponding
Angles Postulate
ABC =
DEF
AAS
• Given: DE =FG ; DA ll
EC; <B and <E are right
angles
• Prove: ABC = DEF
A
C
B
D
E
F
G
Side Angle Side
• Given: AB = BC, AD = EC
• Prove: ABE = CBD
Statements Reasons
AB = BC;
AD = EC
Given
AB = CB
Segment
Addition
<B = <B
Reflexive
Property
ABE =
___CBD
SAS
B
D
E
F
A
C
Hypotenuse-Leg
Statements
Reasons
<1 and <2 are Given
right angles;
AB = CB
<1 = 90°
<2 = 90°
Definition
right angle
<1 = <2
Transitive
Property
BD = BD
Reflexive
Property
HL
ADB =
___CDB
• Given: <1 and <2 are
right angles; AB = CB
• Prove: ADB = CDB
D
1
A
2
B
C
Side Side Side Theorem
Statements
Reasons
<1= <2, <3= <4
Given
BF=BF
Reflexive
ABF=
CBF
ASA
AB=BC
CPCTC
AF=CF
CPCTC
ABC is
isosceles
Def of isosceles
BD – angle
bisector
Def- angle
bisector
BDperpendicular
bisector
Angle bisector of the
vertex angle of an isos.
triangle is a perpendicular
bisector of the base
AD=CD
Def of perpendicular
bisector
FD=FD
reflexive
AFD=
CFD
SSS
• Given: <1= <2, <3=
<4
• Prove: AFD= CFD
B
F
A
D
C
Base Angle Theorem
Statements
AC=BC
DC- angle
bisector
<ACD=<BCD
CD=CD
ACD=
BCD
<A=<B
proofs
Reasons
Given
construction
• Given: AC=BC
• Prove: <A=<B
Def- angle
bisector
Reflexive
SAS
C
A
CPCTC
D
B
A square is a rhombus
Theorem
Statements
Reasons
• Given: ABCD is a
square Prove: ABCD
is a rhombus
ABCD is a square Given
AB=BC=CD=DA
Definition of a
square
ABCD is a
rhombus
Definition of a
Rhombus
A
D
B
C
If the diagonals of a quadrilateral bisect each other, then
the quadrilateral is a parallelogram
Statements
Reasons
BD bisects AC
Given
BE=ED, AE=EC
Definition of a
bisector
<AEB=<DEC
Vertical angles
AEB= CED,
AED=
CEB
SAS
<ECD=<EAB,
<ECB=<EAD
CPCTC
A
B
E
D
AB parallel to CD, Converse of alt.
BC parallel to AD int. angles
ABCD is a
parallelogram
• Given: BD bisects AC
• Prove: ABCD is a
parallelogram.
definition
C
If one pair of adjacent sides of a parallelogram are
congruent, then the parallelogram is a rhombus.
Statements
Reasons
ABCD is a
parallelogram,
AB=BC
Given
AB=CD, BC=AD
Opposite sides of
a parallelogram
are congruent
CD=AB=BC=AD
transitive
ABCD is a
rhombus
definition
• Given: ABCD is a
parallelogram,
AB=BC
• Prove: ABCD is a
rhombus.
A
D
B
C
If the diagonals of a parallelogram bisect the angles of the
parallelogram, then the parallelogram is a rhombus.
Statements
Reasons
ABCD is a
parallelogram, BD
bisects <ADC and
<ABC, AC bisects
<BAD and <BCD
Given
<BAE=<DAE,
<ADE=<CDE,
<ABE=<CBE,
<BCE=<DCE
Definition of an angle
bisector
<EAD=<ECB
Alt. int. angles
<EAB=<ECB
transitive
BE=BE
reflexive
ABE=
CBE
AAS
AB=BC
CPCTC
ABCD is a rhombus
1 pair of sides of a
parallelogram are
congruent.
• Given: ABCD is a
parallelogram, BD
bisects <ADC and
<ABC, AC bisects
<BAD and <BCD.
• Prove: ABCD is a
rhombus.
B
C
E
A
D
If the diagonals of a parallelogram are perpendicular, then
the parallelogram is a rhombus:
Statements
Reasons
ABCD is a
parallelogram, BD
perpendicular to AC.
Given
BE=BE
reflexive
• Given: ABCD is a
parallelogram, BD
perpendicular to AC.
• Prove: ABCD is a
rhombus.
B
AE=EC
Diagonals of a
parallelogram bisect
<BEC=90,
<BEA=90
Definition of
perpendicular
<BEC=<BEA
transitive
ABE=
CBE
A
E
SAS
AB=BC
CPCTC
ABCD is a rhombus
1 pair of adjacent sides of
a parallelogram are
congruent.
D
C