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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.

Grade:10

2 Answers

Aditya Gupta
2081 Points
4 years ago
dear student, first draw the diagram according to the question.
now, let angle ZWP= angle PWX= θ
also, angle ZYQ= angle XYQ= p
cyclic implies ZWX + XYZ = 180
2θ + 2p= 180
or θ + p= 90
now, join PY. and note than ZWP= PYZ= θ (angles in the same segment theorem)
so, PYQ= PYZ + ZYQ
= θ + p= 90
or 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 :)
Vikas TU
14149 Points
4 years ago
Dear student 
Sum of opposite angles of a cyclic quadrilateral being 
180
∘
 we have for cyclic quadrilateral
Angle 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 

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