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Answers for the lesson “Show that a
Quadrilateral is a Parallelogram”
LESSON
8.3
12.
Skill Practice
y
B
1. The definition of a parallelogram
is that it is a quadrilateral with
opposite pairs of parallel sides.
Since }
AB, }
CD, and }
AD, }
BC are
opposite pairs of parallel sides,
quadrilateral ABCD is a
parallelogram.
1
21
A
C
2. Yes; both pairs of opposite sides
D
are congruent.
Sample answer: AB}
5 CD 5 4
and BC 5 DA 5 Ï61
3. The congruent sides must be
opposite one another.
x
13.
y
B
C
4. Theorem 8.8
5. Theorem 8.7
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6. Theorem 8.10
A
7. Since both pairs of opposite
sides of JKLM always remain
congruent, JKLM is always a
parallelogram and }
JK remains
}
parallel to ML.
8. 4
2
10. }
3
9. 8
11.
D
1
22
x
Sample answer: AB}
5 CD 5 5
and BC 5 DA 5 Ï65
14.
B y
y
C
B
1
A
2
A
22
21
C
x
D
x
Sample answer: AB 5 CD 5 5
and BC 5 DA 5 8
D
}
Sample answer: AB 5 CD 5 Ï 41
and BC 5 DA 5 5
Geometry
Answer Transparencies for Checking Homework
233
15. Sample answer: Show
nADB > nCBD using the
SAS Congruence Postulate. This
makes }
AD > }
CB and }
BA > }
CD
using corresponding parts of
congruent triangles
are congruent.
16. Sample answer: Show
nADB > nCBD using the ASA
Congruence Theorem. This makes
} > CD
} and AD
} > CB
} using
AB
corresponding parts of congruent
triangles are congruent.
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
17. Sample answer: Show }
AB i }
DC
by the Alternate Interior Angles
Converse, and show }
AD i }
BC by
the Corresponding Angles
Converse.
18. A
19. 114
20. 45
21. 50
22. A quadrilateral is a parallelogram
if and only if both pairs of
opposite sides are congruent.
23. A quadrilateral is a parallelogram
if and only if both pairs of
opposite angles are congruent.
25. (23, 2); since }
DA must be
parallel and congruent to }
BC, use
}
the slope and length of BC to find
point D by starting at point A.
26. (3, 1); since }
AB must be
parallel and congruent to }
CD, use
}
the slope and length of AB to find
point D by starting at point C.
27. (25, 23); since }
DA must be
parallel and congruent to }
BC, use
}
the slope and length of BC to find
point D by starting at point A.
28. (7, 22); since }
AB must be
parallel and congruent to }
CD, use
}
the slope and length of AB to find
point D starting at point C.
29. Sample answer: Draw a line
passing through points A and B.
]›
At points A and B, construct AP
]› such that the angle each
and BQ
ray makes with the line is the
same. Mark off congruent
segments starting at points A and
]› and BQ
]›, respectively.
B along AP
Draw the line segment joining
these two endpoints.
24. Joining the endpoints of the two
line segments that bisect one
another forms a quadrilateral.
Using Theorem 8.10 you know
that it is a parallelogram.
Q
P
A
B
Geometry
Answer Transparencies for Checking Homework
234
30. 8; since ABCD is a parallelogram
you know that }
AD > }
CB and
}
}
AD i CB. ŽADE > ŽCBF using
34.
the Alternate Interior Angles
Congruence Theorem. It was
The point of intersection of
given that }
BF > }
DE, therefore
the diagonals is not necessarily
nADE > nCBF by the SAS
their midpoint.
Congruence Theorem. Using
corresponding parts of congruent 35.
triangles are congruent, you know
that CF 5 EA 5 8.
Problem Solving
31. a. EFJK, FGHJ, EGHK; in each
The opposite sides that are not
marked in the given diagram are
not necessarily the same length.
case opposite pairs of sides are
congruent.
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b. Since EGHK is a
parallelogram, opposite sides
are parallel.
36.
8
8
32. AEFD and EBCF are
parallelograms by Theorem 8.8 so
}
AD and }
BC both remain parallel
}
}
to EF; AE and }
DF, }
BE and }
CF.
The sides of length 8 are not
necessarily parallel.
33. 1st column: Alternate Interior
Angles Congruence Theorem;
Reflexive Property of Segment
Congruence; Given
2nd column: SAS; Corr. Parts of
s are >; Theorem 8.7
>n
Geometry
Answer Transparencies for Checking Homework
235
37. In a quadrilateral, if consecutive
angles are supplementary, then the
quadrilateral is a parallelogram. In
ABCD you are given ŽA and
ŽB are supplementary, and ŽC
and ŽB are supplementary, which
gives you mŽA 5 mŽC. Also
ŽB and ŽC are supplementary,
and ŽC and ŽD are
supplementary which gives you
mŽB 5 mŽD. So ABCD is a
parallelogram by Theorem 8.8.
B
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
A
C
D
38. It is given that ŽA > ŽC and
ŽB > ŽD. Let mŽA 5 mŽC 5
x and mŽB 5 mŽD 5 y. Since
ABCD is a quadrilateral, you
know that 2x 1 2y 5 360
using the Polygon Interior Angles
Theorem, so x 1 y 5 180. Using
the definition of supplementary
angles, ŽA and ŽB, ŽB and
ŽC, ŽC and ŽD, and ŽD and
ŽA are supplementary. Using
Theorem 8.5, ABCD is a
parallelogram.
} > MP
} and
39. It is given that KP
} > LP
} by definition of segment
JP
bisector. ŽKPL > ŽMPJ and
ŽKPJ > ŽMPL since they are
vertical angles. nKPL > nMPJ
and nKPJ > nMPL by the
SAS Congruence Postulate.
Using corresponding parts of congruent triangles are congruent,
}
} and JM
} > LK
}. Using
KJ > ML
Theorem 8.7, JKLM is
a parallelogram.
40. It is given that DEBF is a
parallelogram and AE 5 CF.
Since DEBF is a parallelogram,
you know that FD 5 EB,
ŽBFD > ŽDEB, and ED 5 FB.
AE 1 EB 5 CF 1 FD which
implies that AB 5 CD, which
} > CD
}. ŽBFC and
implies that AB
ŽBFD, and ŽDEB and ŽDEA
form linear pairs, thus making
them supplementary. Using the
Congruent Supplements Theorem,
ŽBFC > ŽDEA making
nAED > nCFB using SAS.
Using corresponding parts of
congruent triangles are congruent,
}
}. Theorem 8.7 tells you
AD > CB
that ABCD is a parallelogram.
Geometry
Answer Transparencies for Checking Homework
236
41.
F
B
E
C
2. The sum of the interior angles of
G
3. x 5 4, y 5 4; the diagonals of a
A
H
D
Sample answer: Consider the
} is the midsegment
diagram. FG
of nCBD and therefore is parallel
} and half of its length. EH
}
to BD
is the midsegment of nABD and
}
therefore is parallel to BD
and half of its length. This makes
} both parallel and
} and FG
EH
congruent. Using Theorem 8.9,
EFGH is a parallelogram.
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
ABCDE is 5408. To find mŽA and
mŽC, subtract 2708 from 5408
and divide by 2.
parallelogram bisect each other.
Set 12x 1 1 5 49 and 8y 1 4 5
36 and solve for x and y.
4. 85;
8
5
} is the midsegment of nAED
42. FJ
} and
and therefore is parallel to AD
} is the
half of its length. GH
midsegment of nBEC and
} and
therefore is parallel to BC
half of its length. Together this
} and FJ
}.
} > GH
} i GH
gives you FJ
Using Theorem 8.9, FGHJ is a
parallelogram.
Mixed Review of
Problem Solving
1. a. 5; pentagon
5. In a parallelogram consecutive
interior angles are supplementary.
Solve x 1 3x 2 12 5 180 for x.
Use the value of x to find the
degree measure of the two
consecutive angles. In a
parallelogram the angle opposite
each of these known angles has
the same measure.
b. 5408
c. 3608
Geometry
Answer Transparencies for Checking Homework
237
6. a. EFGH is a parallelogram
by Theorem 8.9. As EFGH
changes the shape, ŽE and
ŽG remain congruent, and
ŽH and ŽF remain
congruent, keeping }
FG i }
EH.
b. mŽE and mŽG decrease from
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
558 to 508. mŽH and mŽF
change from 1258 to 1308.
Since EFGH is always a
parallelogram, mŽF and
mŽH are always the same and
mŽE and mŽG are always the
same. Since mŽE is always
supplementary to the mŽF,
their sum must be 1808.
Alternate Interior Angles
Congruence Theorem,
} > CD
} since
ŽBAX > ŽDCY. AB
opposite sides of a
parallelogram are congruent.
nBXA > nDYC by the AAS
Congruence Theorem, making
}
} using corresponding
BX > DY
parts of congruent triangles are
congruent. Use Theorem 8.9 to
show XBYD is a parallelogram.
7. a. Since the slope of }
MN and }
PQ
3
is }
and the slope of }
NP and
11
5
}
QM is 2}4 , then it is a
parallelogram by definition.
}
b. MN 5 PQ 5 Ï130 making
} and NP 5 QM 5
} > PQ
MN
}
}. Using
} > QM
Ï41 making NP
the Theorem 8.3, MNPQ is
a parallelogram.
} i DY
} using the Lines
8. BX
Perpendicular to a Transversal
} and
} > AC
Theorem. Since BX
}, then ŽBXA and ŽDYC
}> AC
DY
are right angles making
ŽBXA > ŽDYC. Using the
Geometry
Answer Transparencies for Checking Homework
238