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If the non parallel sides of a trapezium are equal, prove that it is cyclic.

this question is very urgently. please give me a very fast answer.plz...........


The Brainliest Answer!

AD = BC     (Given)
DE = CF    (Distance between parallel sides is same)
∠AED = ∠BFC = 90°
ΔAED ≅ ΔBFC  (RHS Congruence criterion)
Hence ∠DAE = ∠CBF  (CPCT)  … (1)
Since AB||CD, AD is transversal
∠DAE + ∠ADC = 180°  (Sum of adjacent interior angles is supplementary)
⇒ ∠CBF + ∠ADC = 180°  [from (1)]
Since sum of opposite angles is supplementary in trapezium ABCD.
Thus ABCD is a cyclic trapezium 

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