Question

prove that if the diagonals of a trapezium are
equal then it is cyclic quadrilateral.​

Answer 1

Answer:

angleADE + angleABC = angleADE + 90 + angleFBC = angleADE + angleEAD + 90 = 90 + 90 = 180 degrees. This is consistent to the theorem that any quadrilateral for which the sum of its angles lying in the ends of one of its diagonals being equal to 180 degrees is cyclic.

Step-by-step explanation:

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Answer 2

Answer:

If a pair of opposite sides of a cyclic quadrilateral are equal, then prove that its diagonals are also equal.

Firstly, prove that ΔACD is congruent to ΔBDC by SAS and then further prove the required result

Given Let ABCD be a cyclic quadrilateral and AD = BC.

Join AC and BD.

To prove AC = BD

Proof In ΔAOD and ΔBOC,

∠OAD = ∠OBC and ∠ODA = ∠OCB

[since, same segments subtends equal angle to the circle]

AB = BC [given]

ΔAOD = ΔBOC [by ASA congruence rule]

Adding is DOC on both sides, we get

ΔAOD+ ΔDOC ≅ ΔBOC + ΔDOC

⇒ ΔADC ≅ ΔBCD

AC = BD [by CPCT]

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