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) AD=BC (given) angle ADO=angle 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 .t)      AO+OC=BO+OD Therefore, diagonal AC= diagonal BD                               Hence, proved    

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