Q28 of 63 Page 156

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.


Construction:


Join AF and FD


Proof:


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.


Hence proved.


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