Given: A trapezium ABCD in which AB || CD and AD = BC To prove: ABCD is a cyclic trapezium. Construction: Draw DE perpendicular AB and CF perpendicular AB In right triangles AED and BFC, we have AD = BC (Given) ∠ DEA = ∠ CFB (Each equal to 90%) and, DE = CF (Distance between two parallel lines) ⇒ △ DEA ≅ △ CFB (RHS congruence criterion) ∠ A = ∠ B (CPCT) ....(i) ∠ ADE = ∠ BCF (CPCT) ...(ii) ⇒ ∠ C = ∠ BCF + 90 ° = ∠ ADE + 90° = ∠ ADC ...(iii) ⇒ ∠ C = ∠ D Now, in quadrilateral ABCD, we have ∠ A + ∠ B + ∠ C + ∠ D = 360° (By Angle sum property) ⇒ 2 ∠ A + 2 ∠C = 360° (From (i) and (iii) ∠ A + ∠ C = 180° Hence, quadrilateral ABCD is cyclic.