Given: ABCD is a trapezium whose two non-parallel sides AB and BC are equal.
To Prove: Trapezium ABCD is a cyclic.
Construction: Draw BE ║ AD.
Proof : ∴ AB ║ DE [Given]
AD ║ BE [By construction]
∴ Quadrilateral ABCD is a parallelogram.
∴ ∠BAD = ∠BED ....(i) [Opp. angles of a║gm]
And, AD = BE ....(ii) [Opp. sides of a ║gm]
But AD = BC ...(iii) [Given]
From (ii) and (iii),
BE = BC
∴ ∠BEC = ∠BCE ....(iv)
[Angles opposite to equal sides]
∠BEC + ∠BED = 1800 [Linear Pair Axiom]
⇒ ∠BCE + ∠BAD = 1800 [From (iv) and (i)]
⇒ Trapezium ABCD is cyclic.
[∴ If a pair of opposite angles of a quadrilateral 180 0 , then the quadrilateral is cyclic]
