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Exercise 10.2 Questions 13 |
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Answer» Or solution h bhai kis ka Similarly, we can prove thatBOC + AOD = 180° 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 = CRDR = DSIn 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 thatBOC + AOD = |
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