if the bisector of the angle p andr of a cyclic quadrilateral pqrs intersect the corresponding circle at a and b respectively then AB is a diameter of the circle .
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Answer:
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Step-by-step explanation:
Given:ABCD is a cyclic quadrilateral.
DP and QB are the bisectors of ∠D and ∠B, respectively.
To prove: PQ is the diameter of a circle.
Proof: Since,ABCD is a cyclic quadrilateral.
∴∠CDA+∠CBA=180
∘
since sum of opposite angles of cyclic quadrilateral is 180
∘
2
1
∠CDA+
2
1
∠CBA=
2
1
×180
∘
=90
∘
⇒∠1+∠2=90
∘
........(1)
But ∠2=∠3 (angles in the same segment QC are equal) ........(2)
⇒∠1+∠3=90
∘
From eqns(1) and (2),
∠PDQ=90
∘
Hence,PQ is a diameter of a circle, because diameter of the circle subtends a right angle at the circumference of the circle.
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Answer:
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Step-by-step explanation:
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