Given: ABCD is a quadrilateral circumscribing a circle whose centre is O. To prove: (i) AOB + COD = (ii) BOC + AOD = Construction: Join OP, OQ, OR and OS. Proof: Since tangents from an external point to a circle are equal. AP = AS, BP = BQ ……….(i) CQ = CR DR = DS In OBP and OBQ, OP = OQ [Radii of the same circle] OB = OB [Common] BP = BQ [From eq. (i)] OPB OBQ [By SSS congruence criterion] [By C.P.C.T.] Similarly, Since, the sum of all the angles round a point is equal to AOB + COD = Similarly, we can prove that BOC + AOD =

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