Question

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

Answer 1

Given: ABCD is a trapezium whose non-parallel sides AD and BC are equal.

To prove: Trapezium ABCD is cyclic.

Join BE, where BE || AD

Proof: Since, AB || DE and AD || BE

Since, the quadrilateral ABED is a parallelogram.

∴ ∠BAD = ∠BED    ...(i)   [Opposite angles of a parallelogram are equal]

And AD = BE        ...(ii)   [Opposite angles of a parallelogram are equal]

But AD = BC    [Given]   ...(iii)

From equations (ii) and (iii),

BC = BE

⇒ ∠BEC = ∠BCE    ...(iv)  [Angles opposite to equal sides are equal]

Also, ∠BEC + ∠BED = 180°   ...[Linear pair axiom]

∴ ∠BCE + ∠BAD = 180°     ...[From equations (i) and (iv)]

If sum of opposite angles of a quadrilateral is 180°, then quadrilateral is cyclic.

Hence, trapezium ABCD is cyclic.

Hence proved.