Prove that the parallelogram circumscribing a circle is a rhombus.
Answers
Step-by-step explanation:
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 in length.
Therefore, AP = AS, BP = BQ, CR = CQ and DR = DS.
Adding the above equations,
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
2AB = 2BC
(Since, ABCD is a parallelogram so AB = DC and AD = BC)
AB = BC
Therefore, AB = BC = DC = AD.
Hence, ABCD is a rhombus.
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Hey Mate :D
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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