Prove that the parallelogram circumscribing a circle is a rhombus

• • NCERT: Text Format • • • • • • • • • • • • • Rationalised NCERT • • • • • • • • • • • • • Old NCERT (2015) • • • • • • • • • • • • • Lab Manuals & Kits • • e-Books for UPSC • • Android App • • NCERT Books • • • • • • • • • H. C. Verma • • • Lakhmir Singh • • • • • • • • • R. D. Sharma • • • • • • • • R. S. Aggarwal • • • • • • • All in One • • • • • • • • • Evergreen Science • • • Together with Science • • • Xam Idea 10 th Science • • Classroom Courses • • • • • • • UPSC Exams • • Teaching • • Banking • • • Hair Accessories • Jewellery • Stationery • Lunch Boxes • • Explore Store Consider a circle circumscribed by a parallelogram ABCD, Let side AB, BC, CD and AD touch circles at P, Q, R and S respectively. To Proof : ABCD is a rhombus. As ABCD is a parallelogram AB = CD and BC = AD …[1] [opposite sides of a parallelogram are equal] Now, As tangents drawn from an external point are equal. We have AP = AS [tangents from point A] BP = BQ [tangents from point B] CR = CQ [tangents from point C] DR = DS [tangents from point D] Add the above equations AP + BP + CR + DR = AS + BQ + CQ + DS AB + CD = AS + DS + BQ + CQ AB + CD = AD + BC AB + AB = BC + BC [From 1] AB = BC …[2] From [1] and [2] AB = BC = CD = AD And we know, A parallelogram with all sides equal is a rhombus So, ABCD is a rhombus. Hence Proved.

Given ABCD be a parallelogram circumscribing a circle with centre O. To Prove : ABCD is a rhombus. We know that the tangents drawn to a circle from an exterior point are equal is length. ∴ AP = AS, BP = BQ, CR = CQ and DR = DS. AP+BP+CR+DR = AS+BQ+CQ+DS (AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ) ∴ AB+CD=AD+BC or 2AB=2AD (since AB=DC and AD=BC of parallelogram ABCD) ∴ AB=BC=DC=AD Therefore, ABCD is a rhombus.

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Let’s assume that A B C D is a rhombus which is inscribed in a circle. So, we have ∠ B A D = ∠ B C D [Opposite angles of a rhombus are equal] And ∠ B A D + ∠ B C D = 1 8 0 ∘ [Pair of opposite angles in a cyclic quadrilateral are supplementary] So, 2 ∠ B A D = 1 8 0 ∘ Thus, ∠ B A D = ∠ B C D = 9 0 ∘ Similarly, the remaining two angles are 9 0 ∘ each and all the sides are equal. Therefore, A B C D is a square. Hence Proved.

Given that, Parallelogram ABCD circumscribing circle C 1. Therefore, AB, BC, CD and AD are tangents to circle C 1. We know that lengths of tangents to a circle from a fixed outer point are equal. ∴ AP = AS … (1) (AP & AS are tangents to circle from point A.) BP = BQ … (2) (BP & BQ are tangents to circle from point B.) CR = CQ … (3) (CQ & CR are tangents to circle form point C.) DR = DS … (4) (DR & DS are tangents to circle from point D.) Now, Adding equation (1), (2), (3) and (4), we get (AP + BP) + (CR + DR) = AS + BQ + CQ + DS = (AS + DS ) + (BQ + CQ). ⇒AB + CD = AD + BC. (∴ AP + BP = AB, AS + DS = AD, BQ + CR = BC & CR + DR = CD) ⇒ AB + AB = AD + AD. (∴ AB = CD & AD = BC because ABCD is a parallelogram) ⇒ 2AB = 2AD ⇒ AB = AD ⇒ CD = AB = AD = BC (∵ AB = CD & BC = AD) Hence, AB = BC = CD = AD So, ABCD is a parallelogram with all equal sides. ∴ ABCD is a rhombus. Hence proved.