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Prove that parallelogram circumscribing a circle is a rhombus |
| Answer» Given ABCD is a ||gm\xa0such that its sides touch a circle with centre O.∴ AB = CD and AB || CD,AD = BC and AD || BCNow, P, Q, R and S are the touching point of both the circle and the ||gmWe know that, tangents to a circle from an exterior point are equal in length.∴ AP = AS [Tangents from point A] ... (1)\xa0BP = BQ [Tangents from point B]\xa0... (2)\xa0CR = CQ [Tangents from point C]\xa0... (3)\xa0DR = DS [Tangents from point D]\xa0... (4)On adding (1), (2), (3) and (4), we getAP + BP + CR + DR = AS + BQ + CQ + DS⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)⇒ AB + CD = AD + BC⇒ AB + AB = BC + BC [∵ ABCD is a ||gm .\xa0∴ AB = CD and AD = BC]⇒ 2AB = 2BC⇒ AB = BCTherefore, AB = BC impliesAB = BC = CD = ADHence, ABCD is a rhombus.\xa0In rhombus, it is not necessary that diagonals are equal. If they are equal, then rhombus is considered as a square whose diagonals are always equal. So, there isn\'t any use of proving that the diagonals of a rhombus are equal. | |