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Question

If APB and CQD are two parallel lines, then the bisectors of APQ,BPQ,CQP and PQD 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
APB+BPQ=180o ...angles in linear pair
APE+QPE+BPF+FPQ=180o
2EPQ+2FPQ=180o
[Since PE and PF are bisectors of APQ and BPQ respectively]
Dividing by 2 on both sides we get,
EPQ+FPQ=90o
EPF=90o ...angle addition property ...(I)
Similarly, we can prove EQF=90o ...(II)

Now, APBCQD ...(given)
APQPQD ...alternate angles
and BPQCQP ...alternate angles

12APQ=12PQD and 12BPQ=12CQP

EPQ=FQP and EQP=FPQ ...(III)


In EPQ and FPQ,
EPQFQP ...from (III)
EQPFPQ ...from (III)
side PQ side PQ ... common side
EPQFPQ ...by ASA test
EF ....C.P.C.T ....(IV)

In EPFQ
E+EPF+F+EQF=360o ...Angle sum property of quadrilateral
2E+90o+90o=360o ...from (I), (II), (IV)
2E=180o
E=90o
E=F=90o ...from (IV) ...(V)

EPFQ is a rectangle ...from (I), (II) (V)
[By definition of rectangle]

422501_77838_ans_fc41329677a74858aba81efd36b9c97f.png

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