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If bisectors of opposite angles of a cyc...

If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.

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Given, ABCD is a cyclic quadrilateral.
DP and QB are the bisectors of ` angleD and angleB`, respectively.
To prove PQ is the diameter of a circle.
Construction Join QD and QC.

Proof Since, ABCD is a cyclic quadrilateral.
`:. angle CDA+ angleCBA = 180^(@)`
[sum of opposite angles of cyclic quadrilateral is `180^(@)`]
On dividing both sides by 2, we get
`1/2angleCDA+1/2angleCBA=1/2 xx 180^(@)=90^(@)`
`rArr angle1+angle2=90^(@)`...(i)
`[angle1=1/2angleCDA and angle2=1/2angleCBA]`
But ` angle2=angle3` [ angles in the same segment QC are equal] ...(ii)
`angle1+angle3=90^(@)`
From Eqs. (i) and (ii), `anglePDQ=90^(@)`
Hence, PQ is a diameter of a circle, because diameter of the circle.
Subtends a right angle at the circumference.
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