hi i m jiya had a dout
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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
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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souravsarkar045:
thanks for making the answer as brainliest
wlc
okk dear
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