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Question
The bisectors of the opposite angles A and C of a cydic quadrilateral ABCD intersect the cirde at the points E and F, respectively. Prove that EF is a diameter of the circle.
Solution
In cyclic quadrilateral ABCD
∠A + ∠ C = 180°
1/2 ∠A + 1/2 ∠C = 90°
∠EAB + ∠BCF = 90° -(1) (AE bisects ∠ A ; CF bisects ∠C)
Also ,
∠BCF = ∠BAF - (2) (Angles in the same segment)
Using (1) in (2) we get ,
∠EAB + ∠BAF = 90°
∠FAE = 90°
EF is the diameter of the circle ,
∴ angle in a semi circle is a right angle.
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