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Byju's Answer
Standard IX
Mathematics
Rectangle
If APB and CQ...
Question
If APB and CQD are two parallel lines, then the bisectors of
∠
A
P
Q
,
∠
B
P
Q
,
∠
C
Q
P
and
∠
P
Q
D
forms
A
a square
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B
a rhombus
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C
a rectangle
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D
any other parallelogram
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Solution
The correct option is
C
a rectangle
∠
A
P
B
+
∠
B
P
Q
=
180
o
...angles in linear pair
∴
∠
A
P
E
+
∠
Q
P
E
+
∠
B
P
F
+
∠
F
P
Q
=
180
o
∴
2
∠
E
P
Q
+
2
∠
F
P
Q
=
180
o
[Since PE and PF are bisectors of
∠
A
P
Q
and
∠
B
P
Q
respectively]
Dividing by 2 on both sides we get,
∠
E
P
Q
+
∠
F
P
Q
=
90
o
∴
∠
E
P
F
=
90
o
...angle addition property ...(I)
Similarly, we can prove
∠
E
Q
F
=
90
o
...(II)
Now,
A
P
B
∥
C
Q
D
...(given)
∴
∠
A
P
Q
≅
∠
P
Q
D
...alternate angles
and
∠
B
P
Q
≅
∠
C
Q
P
...alternate angles
∴
1
2
∠
A
P
Q
=
1
2
∠
P
Q
D
and
1
2
∠
B
P
Q
=
1
2
∠
C
Q
P
∴
∠
E
P
Q
=
∠
F
Q
P
and
∠
E
Q
P
=
∠
F
P
Q
...(III)
In
△
E
P
Q
and
△
F
P
Q
,
∴
∠
E
P
Q
≅
∠
F
Q
P
...from (III)
∠
E
Q
P
≅
∠
F
P
Q
...from (III)
side
P
Q
≅
side
P
Q
... common side
∴
△
E
P
Q
≅
△
F
P
Q
...by
A
S
A
test
∴
∠
E
≅
∠
F
....C.P.C.T ....(IV)
In
□
E
P
F
Q
∠
E
+
∠
E
P
F
+
∠
F
+
∠
E
Q
F
=
360
o
...Angle sum property of quadrilateral
∴
2
∠
E
+
90
o
+
90
o
=
360
o
...from (I), (II), (IV)
∴
2
∠
E
=
180
o
∴
∠
E
=
90
o
∴
∠
E
=
∠
F
=
90
o
...from (IV) ...(V)
∴
□
E
P
F
Q
is a rectangle ...from (I), (II) (V)
[By definition of rectangle]
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Similar questions
Q.
If APB and CQD are two parallel lines, then the bisectors of
∠APQ, ∠BPQ, ∠CQP and ∠PQD enclose a
(a) square
(b) rhombus
(c) rectangle
(d) kite
Q.
If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form a ____________.