Question

Prove that a cyclic trapezium is isosceles and it's diagonals are equal

Answer 1

Let ABCD be the cyclic trapezium in which AB||CD.

 We have to prove that AD = BC

 Since AB||CD and BD is the transversal, we have

 ∠ABD =�∠BDC [Alternate angles]

 Also chord AD subtends�∠ABD and chord BC subtends�∠BDC on the circle at B and D respectively.

 Also�∠ABD =�∠BDC (proved above)

 Therefore, AD = BC

Answer 2

According to question

If ABCD is the trapezium which is cyclic in nature

The cyclic nature of the trapezium means that it can be inscribed in a circle
Thus

Clearly     AB =  CD
Thus Making it isosceles in nature

Now when AC and BD intersect each other the angle formed at the center is 90 deg

Thus     Clearly AC = BD