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Prove that the paralleogram circumscribing a circle is a rhombus |
| Answer» Given:-a circle with center O.a perellelogram ABCD touching the circle at points P, Q, R and S.To prove:- ABCD is rhombus.Proof :- a rhombus is a perellelogram with all sides equal.So, we have to prove all sides equal. In perellelogram ABCD,AB=CD & AD=BC, ( opposite sides of perellelogram are equal.)..................(01)From theorem 10.2, length of tangent drown from external point of circle are equal. Hence, AP = AS,........... (02)BP = BQ,............(03)CR = CQ, ............(04)DR = DS. ............(05)Adding equation (02) + (03) + (04) + (05) AP + BP + CR + DR = AS + BQ + CQ + DS (AP + BP)+(CR + DR) = (AS + DS)+(BQ + CQ) AB + CD = AD + BCBut from equation (01) : AB=CD & AD=BC AB + AB = AD + AD 2AB = 2AD AB = ADSo, AB = AD & AB=CD & AD=BC.So, AB = CD = AD = BC.So, ABCD is a perellelogram with all sides equal,There fore. ABCD is rhombus.Hence proved. | |