Prove cyclic trapezium is isosceles and its diagonals are equal to each other.

Solution :



Let ABCD be a cyclic trapezium in which AD || BC. Let AC and BD are the diagonals of the cyclic trapezium ABCD.

In a cyclic trapezium ABCD ,

∠BAD + ∠BCD = 180°   [Opposite angles of cyclic quadrilateral are supplementary]      .......(1)

Since, AD || BC and AB is a transversal, then 

∠BAD + ∠ABC = 180°     [Consecutive interior angles on same side of transversal are supplementary]        ........(2)

From (1) and (2), we have

∴ ∠ABC = ∠BCD

In ∆ABC and ∆DBC,

∠ABC = ∠BCD    (proved above)

∠BAC = ∠CDB (angles in the same segment)

BC = BC (common side)

∴ ∆ABC ≅ ∆BDC     (AAS)

∴ AB = CD and AC = BD   (CPCT)

Hence, cyclic trapezium ABCD is isosceles and diagonals are equal to each other.