asa- angle side angle used to prove triangle congruence: if two angles and the included side of two...
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ASA- Angle Side AngleUsed to prove triangle congruence: if two angles and the included side of two triangles are congruent, then the triangles are congruent.
M
ON
LK Given: KL and NO are parallel; M bisects KO.
Prove: KLM ≡ ONM
Statements Reasons
KL and NO are parallel; M bisects KO.
KML ≡ OMN
∟MKL ≡ ∟MON
KM ≡ MO
KLM ≡ ONM
Given
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
E F GD
Side Angle Side
Statements Reasons
AB = BC; AD = EC
Given
AB = CB Segment Addition
<B = <B Reflexive Property
ABE = ___CBD
SAS
• Given: AB = BC, AD = EC• Prove: ABE = CBD
A C
F
D E
B
Hypotenuse-Leg
Statements Reasons
<1 and <2 are right angles; AB = CB
Given
<1 = 90°<2 = 90°
Definition right angle
<1 = <2 Transitive Property
BD = BD Reflexive Property
ADB = ___CDB
HL
• Given: <1 and <2 are right angles; AB = CB
• Prove: ADB = CDB
A CB
D
1 2
Side Side Side Theorem
• Given: <1= <2, <3= <4
• Prove: AFD= CFD
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
BD- perpendicular 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
A D C
B
F
Base Angle Theorem
• Given: AC=BC• Prove: <A=<B
Statements Reasons
AC=BC Given
DC- angle bisector
construction
<ACD=<BCD Def- angle bisector
CD=CD Reflexive
ACD= BCD
SAS
<A=<B CPCTC
A D B
C
proofs
A square is a rhombusTheorem
Statements Reasons
ABCD is a square Given
AB=BC=CD=DA Definition of a square
ABCD is a rhombus
Definition of a Rhombus
• Given: ABCD is a square Prove: ABCD is a rhombus
A B
D 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
AB parallel to CD,BC parallel to AD
Converse of alt. int. angles
ABCD is a parallelogram
definition
• Given: BD bisects AC• Prove: ABCD is a
parallelogram.
A B
D C
E
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 B
D 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.
A D
B C
E
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
AE=EC Diagonals of a parallelogram bisect
<BEC=90, <BEA=90
Definition of perpendicular
<BEC=<BEA transitive
ABE= CBE SAS
AB=BC CPCTCABCD is a rhombus 1 pair of adjacent sides of
a parallelogram are congruent.
• Given: ABCD is a parallelogram, BD perpendicular to AC.
• Prove: ABCD is a rhombus.
A C
B
D
E