Q4 of 33 Page 57

Prove that a non-isosceles trapezium is not cyclic.

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



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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