Prove that in a cyclic quadrilateral, sum of opposite sides is equal


Since length of tangents drawn from an external point to a circle are equal.
DR=DS
CR=CQ
AP=AS
BP=BQ
Adding above 4 equations, we get
DR+CR+AP+BP = DS+CQ+AS+BQ
( DR+CR ) + ( AP+BP ) = ( DS+AS ) + ( CQ+BQ
DC+AB=DA+CB

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