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.