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Ridge Community High School *
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ZZV9999
Subject
Mathematics
Date
Nov 24, 2024
Type
Pages
4
Uploaded by ConstableHornetMaster75
Name
Date
Class,
Reteach
i
Geometric
Proof
of
a
conditional.
Apply
deductive
reasoning.
To
write
a
geometric
proof,
start
with
the
hypothesis
Prove
that
the
conclusion
of
the
conditional
is
true.
Hypothesis
Deductive
Reasoning
+
Definitions
«
Properties
»
Postulates
*
Theorems
Conclusion
ZABD
=
£1, then
£DBC
=
£1.
Conditional:
If
BD
is
the
angle
bisector
of
ZABC,
and
Prove:
£DBC
=
£1
|
Given:
BD
is
the
angle
bisector
of
ZABC,
and
ZABD
=
£1.
VA
Proof:
1.
BD
is
the
angle
bisector
of
ZABC.
1.
Given
2.
ZABD
=
«DBC
2.
Def.
of
£
bisector
C
3.
LABD
=
£1
3.
Given
’
4.
ZDBC=
/1
4.
Transitive
Prop.
of
=
1.
Given:
Nis
the
midpoint
of
M_iQis_the
midpoint
of
RP,
and
PQ=NM
..
Prove:
PN=QR
Write
a
justification
for
each
step.
Proof:
,
1.
Nis
the
midpoint
of
MP
.
2.
Qis
the
midpoint
of
RP
.
PN
=
NM
PQ=NM
PN=PQ
PQ=0QR
PN=QR
Oult}exv\./
da
Oir
Mfio:nl—-
\gwe/w
Transihue
Pro
&
Sashtubs
OJE%
iria
NUAPoINE
TrinSthue
prop
&
.
&2
1
2
3
4
5
6
7
SUBSTITUTR
Original
content
Copyright
@
by
Holt
McDougal.
Additions
and
changes
to
the
original
content
are the
responsibility
of
the
instructor.
2-46
Holt
Geometry
Name
Date
Class,
Reteach
L2l
Geometric
Proof
continued
A
theorem
is
any
statement
that
you
can prove.
You
can
use
two-column
proofs
and
deductive
reasoning
to
prove
theorems.
Theorem
congruent
angles),
then
the
two
angles
are
congruent.
Congruent
Supplements
|
if
two
angles
are
supplementary
to
the
same
angle
(or
to
two
Right
Angle
Congruence
|
All
right
angles
are
congruent.
Theorem
Here
is
a
two-column-proof
of
one
case
of
the
Congruent
Supplements
Theorem.
Given:
Z4
and
Z5
are
supplementary
and
A
£5
and
£8
are
supplementary.
PR
Prove:
Z4
=6
Proof:
Statements
‘
Reasons
1.
Z4
and
£5
are
supplementary.
1.
Given
2.
/5
and
£6
are
supplementary.
2.
Given
3.
m£4
+
m«L5
=180°
3.
Definition
of
supplementary
angles
4.
ms5
+
m«6
=
180°
4.
Definition
of
supplementary
angles
5.
mZ4
+
ms5
=ms5
+
ms6
5.
Substitution
Property
of
Equality
6.
mcd=ms6
6.
Subtraction
Property
of
Equality
7.
L4=
26
|
7.
Definition
of
congruent
angles
Fill
in
the
blanks
to
complete
the
two-column
proof
of
the
Right
Angle
Congruence
Theorem.
-
¥
l
2.
Given:
£1
and
£2
are
right
angles.
Fz
Prove:
Z1=
/2
Proof:
Statements
-
.
Reasons
1.0
Al
LR
Al
rE
LS,
1.
Given
2.
mz1=90°
2.b.
_cepayre.
L.
3c_
MLR=40°
3.
Definition
of
right
angte
°
4.
mat
=me2
4.d.
_Suhstrtuhon
pop
=
5.6
A
e
5.
Definition
of
congruent
angles
Original
content
Copyright
©
by
Holt
MeDougal.
Additions
and
changes
to
the
original
content
are
the
responsibility
of
the
instructor.
2-47
Holt
Geometry
O
Name
Date
3
e
Practice
A
Geometric
Proof
(Use
one
justification
twice.)
)
o
Write
the
letter
of
the
correct
justification
next
to
each
step.
Given:
HdJ
is
the
bisector
of
Z/HK
and
£1
=
£3.
K
.
b
1.
HJ
is
the
bisector
of
ZIHK.
tfi
A.
Definition
of
£
bisector
i
;
2,
/2=21___
B.
Given
)
3
s1=23_"
C.
Transitive
Prop.
of
=
'
a
s2=s3_C
o
5.Ina
:
am{m
proof,
each
step
in
the
proof
is
on
the
left
and
the
reason
for
the
step
is
on
the
right.
.
-
'
.Fill
in
the
blanks
with
the
justifications
and
steps
listed
to
complete
the
5;
two-column
proof.
Use
this
list
to
complete
the
proof.
.
j}
L1=2L2
.
i
Def.
of
straight
£
'
i
21
and
£2
are
straight
angles.
i
6.
Given:
21
and
42
are
straight
angles.
Prove:
£1
=
/2
Proof:
)
Statements
Reasons
1o
L]
o
LD
Gre
Stight
LS
'
2.
ms1
=
180°,
m£2
=
180°
'
3.
msl1=msL2
Ai
s
LD
1.
Given
2.b.
el
opa.
Straight
£
3.
Subst.
Prop.
of
=
4.
Def.of=
4
4.c.
7.
Given:
21
and
22
form
a
linear
palir,
and
'
£3
and
£4
form
a
linear
pair.
Prove:
mz1
+
m£2
+
m£3
+
mi4
=
360°
Plan:
The
Linear
Pair
Theorem
shows
that
~£1-and
~2
are
supplementary
and
/3
and
24
are
supplementary.
The
definition
of
supplementary
says
that
izl
+
ms2
=
180°
and
m£3
+
m£4
=
180°
Use
the
Addition
Property
i
of
Equality
to
make
the
conclusion.
T
,
e
;
Statements
1.
21
and
£2
form
a
linear
pair,
and
1
/3
and
24
form
a
linear
pair.
-
2.
21
and
22
are
supplementary,.and
/3
and
24
are
supplementary.
.
¥y
e
mef
+mea
=
[80°
mcztEmédd
=|80°
4.ms1
+
ms2
+
mL3
+
mLd
=
360°
Reasons
.a.
C“UW
L4
-
e
-
2
b,
Linecr
Pouy
Theovem
'
3
Def.
of‘supp,'é
'
B
s
a.
Bdd
PJ@P@#
=
'
43
Gopyright
@
by
Holt,
Rinehart
and
Winston.
Holt
Geometry
All
rights
resefved.
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Name
P
;
Date
Class
_
Practice
B
P
Geometric
Proof
Write
a
justification
for
each
step.
£
.
.
-~
N
E
Given:
AB=
EF.
Bis
the
midpoint
of
AC,
2—""
‘
and
£
is
the
midpoint
of
DF.
)
I
.
—_—
c
1.
Bis
the
midpoint
of
AC,
6""
.
and
E
is
the
midpoini
of
DF.
.
:
e
,
2.
AB
=
BC,
and
DE
=
EF.
et
of
midpt
3.
AB
=
BC,
and
DE
=
EF.
.
:
U,AJ
Seqments
-
4.
AB
+
BC
=
AC,
and
DE
+
EF
=
DF.
'
_fie&mmf_&dd_flfiul&k’-
5.
2AB
=
AC,
and
2EF
=
DF.
Scbstitufion
’Pf@f’
=
6.
AB=
EF
.
Gluem
‘
7.
2AB
=
2EF
o
Mutt,
Pep
of
=
8.
AC=DF
Sy
tsttution
Pop
=
IR
9.
AC
=
DF
|
D,e_f-_%_"ég%mflffi
Fill
in
the
bianks
to
complete
the
two-column
proof.
10.
Given:
£HKJis
a
straight
angle.
'
'
Kl
bisects
£
HKJ.
Prove:
£
/KJis
a
right
angle.
Proof:
'
Statements
.
Reasons
1.2
LHIKT
(S
Shrawgnt
&
|1
Gven
:
<
i
1
2.
msHKJ
=
180°
2.
b.
Def
Of
Straught
&
3.c.
W
LHKT
3.
Given
4.
LIKJ=
L
IKH
4.
Def.
of
£
bisector
5.
mLIKJ
=
m2IKH
)
5.
Def.
of
=
4
16.d.MLEKH+
md TRT
=
mLHKI6.
2
Add.
Post.
.
7.
2m2
IKJ
=
180°
17.
e.
subst.
(Stepsu
S,
b
)
8.
mLIKJ=90
-
)
8.
Div.
Prop.
of
=
°
9.
£
IKJis
a
right
angle.
et
D’éP
@fi
rt.
L=
.
%
Copytight
©
by
Holt,
Rinehart
and
Winston,
'
44
°
‘Holt
Geometry
All
rights
reserved.