prove that a cyclic trapezium is isosceles and its diagonals are equal....

let ABCD be the cyclic trapezium with AB IICD.

thru' C draw CE parallel to  AD meeting AB in E.

So

AECD is a parallelogram.

so

angle D=angle AEC.... opp angles of a parallelogram are equal....(i)

but

angle D+angle ABC=180... opp angles of a cyclic quadr are supplementary....(ii)

from (i) and (ii)

angle AEC+angle ABC=180

but

angle AEC+angle CEB= 180...linear pair

so

 angle ABC= angle CEB ..(iii)

so

 CE=CB...  sides opp equal angles are equal.(iv)

 but

CE=AD...opp  sides of parallelogram AECD.

from (iv) we get

AD=CB

Thus cyclic quadri ABCD is isoceles.

this proves the first part of the question.

now,

 join AC and BD, the diagonals.

in triangles DAB and CBA,

AD=CB...proved before

AB=AB common

angle ADB= angle ACB.. angles in the same  segment of a circle are equal.here AB is the chord.

so triangles DAB and CBA are congruent....SAS rule.

so

AD=CB... CPCT

hence proved.