Question

ABCD is a cyclic quadrilateral bisector of angle A and angle C meets the circle at P and Q respectively. prove that PQ is a diameter of the circle ​

Answer 1

Stick to the attatchment :)

Answer 2

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.

Answered By

toppr

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