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.

Question

Aditya Gupta · Answer

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 2 θ + 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 · Answer

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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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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