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WXYZ is a cyclic quadrilateral.if the bisectors of angle XWZ and angle XYZ meet the circle at the points P and Q respectively, then prove that PQ is the diameter of the circle.
WXYZ is a cyclic quadrilateral.if the bisectors of angle XWZ and angle XYZ meet the circle at the points P and Q respectively, then prove that PQ is the diameter of the circle.

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11 months ago

```							dear student, first draw the diagram according to the question.now, let angle ZWP= angle PWX= θalso, angle ZYQ= angle XYQ= pcyclic implies ZWX + XYZ = 1802θ + 2p= 180or θ + p= 90now, join PY. and note than ZWP= PYZ= θ (angles in the same segment theorem)so, PYQ= PYZ + ZYQ= θ + p= 90or PYQ= 90.hence PQ is a diameter, since when P and Q are joined to O (centre of circle), angle POQ= 2*pyq= 180. hence, POQ is a straight line, thereby implying that it is the diameter.kindly approve :)
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11 months ago
```							Dear student Sum of opposite angles of a cyclic quadrilateral being 180∘ we have for cyclic quadrilateralAngle XWZ + Angle XYZ = 180 XWA + XYB = 90 XWA = XYA being the angles on same arc AX So XYA + XYB = 90 AYA = 90 This means AB must be diameter of the circle. Good Luck
```
11 months ago
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