Prove that any non-isosceles trapezium is not cyclic

1.	Prove that any non-isosceles trapezium is not cyclic
Answer» $\huge\star{\underline{\mathtt{\red{A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}⋆$ ☁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 { Note:-figure refer to the attachment}

Answer»

$\huge\star{\underline{\mathtt{\red{A}\pink{N}\green{S}\blue{W}\purple{E}\orange{R}}}}⋆$

☁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

{ Note:-figure refer to the attachment}

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