**Given : **A trapezium ABCD in which AB || CD and AD = BC.

**To Prove :** ABCD is a cyclic trapezium.

**Construction :** Draw DE ⊥ AB and CF ⊥ AB.

**Proof :** In ∆DEA and ∆CFB, we have

AD = BC [Given]

∠DEA = ∠CFB = 90° [DE ⊥ AB and CF ⊥ AB]

DE = CF

[Distance between parallel lines remains constant]

∴ ∆DEA ≅ ∆CFB [RHS axiom]

⇒ ∠A = ∠B ...(i) [CPCT]

and, ∠ADE = ∠BCF ..(ii) [CPCT]

Since, ∠ADE = ∠BCF [From (ii)]

⇒ ∠ADE + 90° = ∠BCF + 90°

⇒ ∠ADE + ∠CDE = ∠BCF + ∠DCF

⇒ ∠D = ∠C ..(iii)

[∠ADE + ∠CDE = ∠D, ∠BCF + ∠DCF = ∠C]

∴ ∠A = ∠B and ∠C = ∠D [From (i) and (iii)] (iv)

∠A + ∠B + ∠C + ∠D = 360° [Sum of the angles of a quadrilateral is 360°]

⇒ 2(∠B + ∠D) = 360° [Using (iv)]

⇒ ∠B + ∠D = 180°

⇒ Sum of a pair of opposite angles of quadrilateral ABCD is 180°.

⇒ ABCD is a cyclic trapezium **Proved**