Question

Prove that any non-isosceles trapezium is not cyclic​

Answer 1

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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

Answer 2

Answer:

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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