Prove that the parallelogram circumscribing a circle is a rhombus.
Answers
consider a parallelogram of ABCD
Dr= ds -----> equation 1
cr= cq --------> equation 2
as=ap -------->equation 3
bp= bq---------> equation 4
add equation 1, 2, 3, 4
(Dr+cr)+(bp+ap)=(ds+as)+ (+cq+bq)
dc+ab=ad+bc
hence it is proved
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Your Answer :---
Since ABCD is a parallelogram,
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AB = CD …(1)
BC = AD …(2)
It can be observed that
DR = DS (Tangents on the circle from point D)
CR = CQ (Tangents on the circle from point C)
BP = BQ (Tangents on the circle from point B)
AP = AS (Tangents on the circle from point A)
Adding all these equations, we obtain
DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
On putting the values of equations (1) and (2) in this equation, we obtain
2AB = 2BC
AB = BC …(3)
Comparing equations (1), (2), and (3), we obtain
AB = BC = CD = DA
Hence, ABCD is a rhombus.
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