If the bisectors of the opposite angles of a cyclic quadrilateral ABCD intersects the circumscribed circle of this quadrilateral at points P and Q, then prove that PQ is a diameter of this circle.
Given ABCD is a cyclic quadrilateral. AP and CQ are bisectors of ∠A and ∠C respectively.
We have to prove that PQ is the diameter of the circle.
Join AF and FD
We know that in a cyclic quadrilateral, opposite angles are supplementary.
⇒∠A + ∠C = 180°
⇒ 1/2 ∠A + 1/2 ∠C = 90°
⇒∠EAD + ∠DCF = 90° … (1)
We know that angles in the same segment are equal.
⇒∠DCF = ∠DAF … (2)
From (1) and (2),
⇒∠EAD + ∠DAF = 90°
⇒∠EAF = 90°
∠EAF is the angle in a semicircle.
∴ EF is the diameter of the circle.
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