prove that a parallelogram circumscribing a circle is a rhombus
Answers
Step-by-step explanation:
let ABCD is parallelogram circumscribing a circle
angle ABC+ ADC=180----angel of cyclic q.
also. BAD+BCD=180-----||-
and ABCD is parallelogram
so ABC=ADC=90
also BAD=BCD=90
ABCD is rectangle
we know that
distance of all points form center of circal is same
ABCD is rhombus
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Your Answer :---
Since ABCD is a parallelogram,
[ Plz see attached file also :) ]
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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