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Question
If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.
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Solution
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.
Construction: Join QD and QC.
Proof: Since, ABCD is a cyclic quadrilateral.
∴ ∠CDA + ∠CBA = 180° ...[Sum of opposite angles of cyclic quadrilateral is 180°]
On dividing both sides by 2, we get
`1/2 ∠CDA + 1/2 ∠CBA = 1/2 xx 180^circ = 90^circ`
⇒ ∠1 + ∠2 = 90° ...(i) `[∠1 = 1/2 ∠CDA "and" ∠2 = 1/2 ∠CBA]`
But ∠2 = ∠3 [Angles in the same segment QC are equal] ...(ii)
∠1 + ∠3 = 90°
From equations (i) and (ii),
∠PDQ = 90°
Hence, PQ is a diameter of a circle, because diameter of the circle.
Subtends a right angle at the circumference.
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