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.

Answer 1

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.