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Question

A quadrilateral ABCD in which AB=a,BC=b,CD=c and DA=d is such that one circle can be inscribed in it and another circle circumscribed about it, then cosA is equal to

A
ad+bcadbc
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B
adbcad+bc
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C
ac+bdacbd
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D
acbdac+bd
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Solution

The correct option is B adbcad+bc
Since circle is inscribed in the quadrilateral ABCD then
a+c=b+d ...........(1)
and quadrilateral is cyclic,then
A+C=π or C=πA ..............(2)
Now, in ABD,
cosA=a2+d2(BD)22ad
or BD2=a2+d22adcosA ...................(3)
and in BCD,
cosC=b2+c2(BD)22bc
or BD2=b2+c22bccosC
=b2+c22bccos(πA) (from (2))
BD2=b2+c2+2bccosA ................(4)
From eqns(3) and (4) we get
2bccosA=BD2b2c2
2bccosA=a2+d22adcosAb2c2
cosA=a2+d2b2c22(bc+ad) ..................(5)
Now, from eqn(1),
a+c=b+d
ad=bc
Squaring both sides, we get
(ad)2=(bc)2
a2+d22ad=b2+c22bc
a2+d2b2c2=2ad2bc ...............(6)
Substituting (6) in (5) we get
cosA=2ad2bc2(ad+bc)=adbcad+bc

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