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