Saved Bookmarks
| 1. |
Prove that the parallekogram circumscribing a circle is a rhombus |
|
Answer» Prove by equalling adjacent sides of a parallelogram Given ABCD is a ||gm such 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) BP = BQ [Tangents from point B] ... (2) CR = CQ [Tangents from point C] ... (3) DR = DS [Tangents from point D] ... (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 . ∴ AB = CD and AD = BC]⇒ 2AB = 2BC⇒ AB = BCTherefore, AB = BC impliesAB = BC = CD = ADHence, ABCD is a rhombus |
|