Math (12)_text.pdf

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SEÏ
I12I
LINE
BISECTORS
AND
ANGLE
BISECTORS
line
l
is
called
a
right
bisèctor
of
a
line
its
Rieht
Bisèctor
of
a
Line
Segment:
A
segment
if
t
is
perpendiculai
to
the
line
segment
and
passes
through
Bisèctor,
oPan
Angle:
mid-point.
A
ray
BP
is
called
the
bisèctor
of
ZABC
if
P
is
a
point
in
the
interior
of
the
angle
and
ZABP
=
ZP
BC.
L-f
ElïTOM
Any
équidistant
+
P
point
on
the
right
bisèctor
of
a
line
segment
is
from
its
end
points.
Given:
A
such
that
line
LM
intersects
the
line
segment
AB
at
the
point
C
rti
LM
XÂBandÂCaBCP
is
a
point
on
PÀ
sPB
p
to
the
points
LM.
To
Prove:
Construction:
Join
A
and
B.
Reasons
Mt
Proof
In
AACP
«
—
ABCP
Statements
ÂC
sBC
ZACP
s
ZBCP
PC=
PC
AACP
=
ABCP
PA
s
PB
Given
given
PC
X
AB
,
so
that
each
Z
at
C
=
90°
common
S.A.S.
postulate
(corresponding
triangles)
sides
of
congruent
Hence
Theorem
Any
is
point
équidistant
from
the
end
points
of
a
line
segment
on
the
right
bisèctor
of
it.
Given
AB
is
a
line
segment.
Point
P
is
such
that
PAsPB
The
Point
P
is
on
the
right
bisèctor
of
AB
To
Prove
.
Joint
P
to
C,
the
midpoint
of
AB.
J_
in
AACP
<
—
ABCP
>
Statements
Reasons
Given
PÀ
s P
B
PC
s
PC
«r
Common
Construction
s.s.s
ÀCsBC
AACP
s
ABCP
=
S.S.S
(corresponding
angles
of
congruent
But
Le.,
Also
ZACPsZBCP
mZACP
+
mZBCP=180°
mZACP
=
inZBCP-90°
PCXAB
CÂsCB
PC
is
a
right
bisector
of
AB
.
(i)
(ii)
triangles)
Snpplementary
angles
From
(üi)
(iv)
(i)
and
(ii)
mZACP
=
90°
(proved)
construction
Le.,
the
point
P
is
on
the
right
bisector
of
from
(iii)
and
(iv)
AB.
1.
Prove
that
the
centre
of
a
Circle
witli
centre
Exercise
12.
E
I
circle
is
on
the
right
bisectors
of
each
of
its
chords.
O
is
To
Provè
chords
Centre
of
the
circle
on
right
bisectors
of
each
of
its
Construction
Draw
any chord
AB
Draw
.
PrOof:
OCX
AB
join
O
with
A
and
B.
Reasors
Statements
In
AOAC
<->
AOBC
Common
Each
of
90°
H.S
s
H.S
Corresponding
triangles.
ÔÂ=ÔB
ÔCsÔC
ZACO
s
ZBCO
/.
Radii
of
same
circle
AACO
=
ABCO
AC
s
BC
OC
is
sides
of
the
congruent
the
right
bisector
of
AB
2.
Where
will
be
the
centre
of
a
three
R
circle
passing
through
non-
collinear
points
and
why?
Three
villages
Circle
is
P,
the
locus
of
a
point
which
Q,
R
not
on
the
moves
so
point
will
3.
same
that
its
distance
froin
a
fixed
line.
O
remains
same.
Otherwise
no
circle
To
Prove
Park
R.
is
be
formed.
équidistant
from
P,
Three
villages
P,
Q
and
R
are
not
on
the
same
line.
The
people
of
these
villages
want
to
make
a
Children
Park
at
such
a
place
which
is
équidistant
front
these
three
villages.
After
fïxing
Q
and
Construction
Complété
the
triangle
PQR,
draw
the
right
bisectors
of
the
sides
PQ
and
with
park.
the
place,
of
Children
park,
prove
that
the
QR
P,
cutting
each
other
at
O.
loin
O
Park
is
équidistant
from
the
three
villages.
Q
and
R.
let
O
be
the
P
roof:
Statements
In
Reasons
AOPC
<->
AOQC
Construction
CPsCQ
OC
s
OC
ZI
sZ2
AOCP
sAOCQ
OP
=
OQ
(i)
...
Common
Each
of
90°
S.
A.
S
~
S.A.S
Corresponding
sides
of
congruent
triangles
Similarly
OQ
s
OR
.
...
(ii)
OP
s
OQ
s
ÔR
Th
co
rem
The
Given
right
bisectors
of
the
sides
of
a
triangle
are
concurrent.
A
A
ABC
To
Prove
The
concurrent.
right
bisectors
of
AB,
BC
and
CA
are
Construction
Draw
the
right
bisectors
of
at
the
point
AB
and
BC
which
A,
Proof:
meet
each
other
O.
Join
O
to
B
and
C.
Statements
Reasons
(i)
OAsOB
(Each
point
on
right
bisector
of
a
segment
is
équidistant
from
its
end
points)
ÔBsÔC
OA
=
OC
Point
(ii)
as
in
(iü)
(i)
From
(O
is
(i)
and
(ii)
0
is
on
the
right
bisector
of
(iv)
équidistant
from
A
and
C)
CA.
But
point
construction
0
is
on
the
right
bisector
of
BC
{from
(v)
(iv)
AB
Hence
and
(v)}
and
of
the
right
bisectors
of
the
three sides
of
a
triangle
are
concurrent
at
O.
Note:
(a)
The
The
right
bisectors
of
the
sides
of
an
acute
triangle
intersect
each
other
inside
the
triangle.
(b)
right
bisectors
of
the
sides
of
a
right
triangle
intersect
each
other
on
the
hypoténuse.
(c)
The
right
bisectors
of
the
sides
of
an
obtuse
triangle
intersect
each
_
other
outside
the
triangle.
Yheorenii
Any
équidistant
point
its
on
anus.
the
bisector
of an
angle
is
from
A
point
P
is
on
OM
,
the
bisectors
of
ZAOB.
OB.
PQsPR
i.e.,
P
is
équidistant
from
OA
and
Construction
Draw
PR
1
OA
and
PQ
±
OB
Proof:
.
Statements
In
Reasons
APOQ
<-
ÔP=ÔP
APOR
Common
Construction
Hence
ZPQO
=
ZPRO
ZPOQ
=
ZPOR
APOQ
s
APOR
PQsPR
Any
point
P
Given
S.A.A.
s
S.
A.
A.
(corresponding
triangles)
sides
of
congruent
Théorem
inside
an
angle,
équidistant
from
its
arms,
is
on
the
bisector
of
it.
A,
Any
where
point
lies
inside
ZAOB
such
that
PQsPR,
PQXOB
and
PR
Z
O
A.
is
To
Prove
Construction
Proof:
Point
P
on
the
bisector
of
ZAOB.
Join
P
to
O.
Reasons
Given
Given
H.S.
(right
angles)
Statements
In
Hence
i.e.,
APOQ
< >
APOR
ZPQO
s
ZPRO
PÔ=PO
PQaPR
APOQ
s
APOR
ZPOQ
s
ZPOR
P
is
—
Common
s
H.S.
(corresponding
angles
of
congruent
triangles)
on
the
bisector
of
ZAOB.

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