Let the cyclic quadrilateral be A, B, C and D.
Now join BD and AC.
BD intersect AC at O
Now in triangle ADO
and triangle BCO
angle BDA=angle ACB(angle at same segment are equal)
CBO(angle at same segment are equal)
Therefore, triangle ADO
is congruent to triangle BCO
Now, AO=OB (c .p .c .t)
OC=OD (c .p. c
Therefore, diagonal AC= diagonal BD