Prove that the parallelogram circumscribing a circle is a rhombus.
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Answer:
parallelogram circumscribing a circle is a rhombus
Step-by-step explanation:
Let say ABCD is a parallogram
=> AB = CD & BC = AD
it circumscribe a circle so
let say circle touches AB , BC , CD & AD at
L , M , N & O respectively
then AL = AO
& BL = BM
CN = CM
DN = DO
AB = AL + BL = AO + BM
CD = CN + DN = CM + DO
=> AB + CD = AO + BM + CM + DO
=> AB + CD = AO + DO + BM + CM
=> AB + CD = AD + BC
AB = CD & BC = AD
=> 2AB = 2BC
=> AB = BC
=> AB = BC = CD = DA
=> Parallelogram is a rhombus ( as rhombus has all 4 sides Equal)
Hence parallelogram circumscribing a circle is a rhombus
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