Answers

2016-03-21T20:24:50+05:30
In ΔAED and ΔBFC,
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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but the proved proprty is crrct
yaa
hmm ..
every where , this proof is given
plz mark brainlest
2016-03-21T20:34:32+05:30
Given :ABCD is an isosceles trapezium 

Construction : Draw EB parallel to AD

To Prove : ABCD is cyclic 

Proof : ABED is  a Parallelogram 

∠A = ∠E 

ΔBEC is an isosceles triangle

∠DEB + ∠BEC = 180°

∠DEB+∠BCE = 180°

∴∠A + ∠C = 180°
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