Prove that any non isosceles trapezium is not cyclic
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☁We will prove by negation
☁Let ABCD be the cyclic trapezium with AB ∥ CD
☁Through C draw CE parallel to AD meeting AB in E
☁Thus, AECD is a parallelogram
parallelogram are equal) …(1)
☁Thus, ∠ D = ∠AEC ( opp. Angle of parallelogram are equal) …(1)
☁But, ∠ D + ∠ ABC = 180° (opp. Angle of a cyclic quadrilateral are
☁Supplementary) ….(2)
☁From (1) and (2)
☁∠ AEC + ∠ ABC = 180°
☁But, ∠ AEC + ∠ CEB = 180° (linear pair)
☁Thus, ∠ AEC + ∠ ABC = ∠ AEC + ∠ CEB
☁⇒ ∠ ABC = ∠ CEB …(3)
☁⇒ CE = CB (side opposite to equal angle are equal) …(4)
☁But, CE = AD (opp. Sides of parallelogram AECD)
☁From (4) we get,
☁AD = CB
☁Thus, cyclic quadrilateral ABCD is isosceles
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