G.CO.C.11 STUDENT NOTES WS #3 – geometrycommoncore.com
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1. Prove that opposite sides of a parallelogram are congruent.
a) Proof by Symmetry and Patty Paper (Informal – Transformational Approach)
Given: Parallelogram ABCD
C
Prove that opposite sides of the parallelogram are congruent.
B
Established in G.CO.3, a parallelogram has rotational symmetry,
when rotated about its center 180.
Rotational symmetry would map AB onto CD and AD onto CB ,
thus making them congruent.
This can be tested using patty paper. Create a parallelogram ABCD
on a piece of paper. Draw in the two diagonals AC and BD to
determine the intersection E. Copy the parallelogram onto the patty
paper. Line up the two parallelograms and then pin the patty paper
with your pencil at E and begin to rotate the patty paper. At a 180
rotation, you will see that AB  CD and AD  CB .
E
D
A
A
D
E
B
C
b) Proof by Congruent Triangles (Formal – Classic Approach)
Given: Parallelogram ABCD
C
Prove: AB  CD , AD  CB
Construct an auxiliary line that is the diagonal BD.
STATEMENT
ABCD is a parallelogram
AB || CD , AD || CB
ABD  CDB
ADB  CBD
BD  BD ,
ABD  CDB
AB  CD , AD  CB 
REASON
Given
Definition of Parallelogram
If ||, then alternate interior ’s 
If ||, then alternate interior ’s 
Reflexive Property (Common Side)
ASA
CPCTC
B
D
A
G.CO.C.11 STUDENT NOTES WS #3 – geometrycommoncore.com
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2. Prove that opposite angles of a parallelogram are congruent.
a) Proof by Symmetry and Patty Paper (Informal – Transformational Approach)
Given: Parallelogram ABCD
C
Prove that opposite angles of the parallelogram are congruent.
B
Established in G.CO.3, a parallelogram has rotational symmetry,
when rotated about its center 180.
Rotational symmetry would map A onto C and B onto D, thus
making them congruent.
This can be tested using patty paper. Create a parallelogram ABCD
on a piece of paper. Draw in the two diagonals AC and BD to
determine the intersection E. Copy the parallelogram onto the patty
paper. Line up the two parallelograms and then pin the patty paper
with your pencil at E and begin to rotate the patty paper. At a 180
rotation, you will see that A C and B  D.
E
D
A
A
D
E
B
C
b) Proof by Congruent Triangles (Formal – Classic Approach)
Given: Parallelogram ABCD
C
Prove: A C, B  D
STATEMENT
REASON
ABCD is a parallelogram
Given
Definition of Parallelogram
AB || CD , AD || CB
ABD  CDB
If ||, then alternate interior ’s 
ADB  CBD
If ||, then alternate interior ’s 
Reflexive Property (Common Side)
BD  BD ,
ASA
ABD  CDB
CPCTC
A C
Using diagonal AC we could follow a similar argument to prove that
B  D
B
D
A
G.CO.C.11 STUDENT NOTES WS #3 – geometrycommoncore.com
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3. Prove that diagonals bisect each other in a parallelogram.
a) Proof by Symmetry and Patty Paper (Informal – Transformational Approach)
Given: Parallelogram ABCD
C
Prove that diagonals bisect each other.
B
E
Established in G.CO.3, a parallelogram has rotational symmetry, when
rotated about its center 180.
D
A
Rotational symmetry would map DE onto BE and CE onto AE , thus
making them congruent.
This can be tested using patty paper. Create a parallelogram ABCD on a
piece of paper. Draw in the two diagonals AC and BD to determine
the intersection E. Copy the CED onto the patty paper. Line up the
patty paper with the original parallelogram and then pin the patty paper
with your pencil at E and begin to rotate the patty paper. At a 180
rotation, you will see that DE  BE and
CE  AE.
b) Proof by Congruent Triangles (Formal – Classic Approach)
Given: Parallelogram ABCD
C
Prove: DE  BE , CE  AE
STATEMENT
ABCD is a parallelogram
AB || CD , AD || CB
CDE  ABE
DCE  BAE
CD  AB
ABE  CDE
DE  BE , CE  AE 
B
REASON
Given
Definition of Parallelogram
If ||, then alternate interior ’s 
If ||, then alternate interior ’s 
Opposite Sides of Parallelogram 
ASA
CPCTC
E
A
D
G.CO.C.11 STUDENT NOTES WS #3 – geometrycommoncore.com
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4. Prove that diagonals are congruent in a rectangle.
a) Proof by Symmetry and Patty Paper (Informal – Transformational Approach)
C
Given: Rectangle ABCD
Prove that diagonals are congruent.
Established in G.CO.3, a rectangle has reflectional symmetry, when
reflected over the line through the midpoints of opposite sides. This
would map B onto C and A onto D, thus the isometric properties of
reflection would preserve the distances of BD and CA .
This can be tested using patty paper. Create a rectangle ABCD on a
piece of paper. Draw in the two diagonals AC and BD . Copy the
rectangle onto the patty paper. Place only one of the diagonals on
the patty paper. Reflect the patty paper by reversing it to the other
side. Match up the two rectangles (A onto D, B onto C, C onto B, D
onto A). Notice that the diagonal on the patty paper matches the
other diagonal. BD  CA .
B
D
A
C
B
D
A
B
C
A
D
b) Proof by Triangle Congruence (Formal – Classic Approach)
C
Given: Rectangle ABCD
B
Prove: BD  AC
STATEMENT
ABCD is a rectangle
A rectangle is a parallelogram
AB  CD
AD  AD
BAD and CDA are right
REASON
Given
Definition of Rectangle
Opposite Sides  in Parallelogram
Reflexive Property (Common Side)
Definition of Rectangle
D
A
G.CO.C.11 STUDENT NOTES WS #3 – geometrycommoncore.com
angles
BAD  CDA
BAD  CDA
BD  AC
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All Right  ' s are 
SAS
CPCTC
5. Prove that the diagonals of a rhombus are angle bisectors.
a) Proof by Symmetry and Patty Paper (Informal – Transformational Approach)
Given: Rhombus ABCD
B
C
Prove that diagonals are angle bisectors.
E
A
Established in G.CO.3, a rhombus has reflectional symmetry, when
reflected over either diagonal. If we use diagonal AC , this would
map B onto D, A onto A and C onto C. Thus the isometric properties
of a reflection would preserve the angle measures. Thus BAC 
DAC and BCA  DCA. Using the other diagonal, we could also
establish that ABD  CBD and ADB  CDB.
D
This can be tested using patty paper. Create a rhombus ABCD on a
piece of patty paper. Draw in the diagonals. Now fold the patty
paper so that A maps onto C, crease the patty paper. Notice that
ABD  CBD and ADB  CDB. Now fold the patty paper so that
B maps onto D, crease the patty paper. Notice that BAC  DAC
and BCA  DCA.
b) Proof by Triangle Congruence (Formal – Classic Approach)
Given: Rhombus ABCD
B
Prove: CA and BD are angle bisectors
STATEMENT
ABCD is a rhombus
AB  BC  CD  AD
A rhombus is a parallelogram
BE  ED
AE  EC
AEB  AED  CEB  CED
REASON
Given
Definition of Rhombus (4  sides)
Definition of Rhombus
Diagonals bisect in parallelogram
Diagonals bisect in parallelogram
SSS
C
A
E
D
G.CO.C.11 STUDENT NOTES WS #3 – geometrycommoncore.com
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EAB  EAD  ECB  ECD CPCTC
Definition of Bisector
CD is an angle bisector
EBA  EBC  EDA  EDC CPCTC
Definition of Bisector
BD is an angle bisector
6. Prove that the diagonals of a rhombus are perpendicular.
a) Proof by Symmetry and Patty Paper (Informal – Transformational Approach)
Given: Rhombus ABCD
B
C
Prove that diagonals are perpendicular.
E
A
Established in G.CO.3, a rhombus has reflectional symmetry, when
reflected over either diagonal. That would require AC to be the
perpendicular bisector of BD . Thus BD  AC .
D
This can be tested using patty paper. Create a rhombus on a piece of
patty paper. Draw in the diagonals. Now fold the patty paper so that
A maps onto C, crease the patty paper. Notice that the crease is the
diagonal. Thus the diagonal is a perpendicular bisector of the other
two vertices making them perpendicular to each other.
b) Proof by Triangle Congruence (Formal – Classic Approach)
Given: Rhombus ABCD
B
C
Prove: BD  AC
A
E
D
STATEMENT
ABCD is a rhombus
A rhombus is a parallelogram
AB  BC  CD  DA
AE  EC
BE  DE
REASON
Given
Definition of a Rhombus
Definition of a Rhombus
Diagonals bisect each other in parallelogram
Diagonals bisect each other in parallelogram
BAE  BCE  DAE  DCE
SSS
G.CO.C.11 STUDENT NOTES WS #3 – geometrycommoncore.com
DEA  DEC  BEA  BEC
mDEA = mDEC = mBEA = mBEC
mDEA + mDEC + mBEA + mBEC =360
mDEA + mDEA + mDEA + mDEA =360
4mDEA =360mDEA =90
BD  AC
CPCTC
Definition of Congruence
Angle Sum about a Point = 360
Substitution Property
Simplify & Division Property
Intersect at Right Angles
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