Answer :

Consider ABCD as a parallelogram touching the circle at points P, Q, R and S as shown

As ABCD is a parallelogram opposites sides are equal

⇒ AB = CD …(a)

⇒ AD = BC …(b)

AP and AS are tangents from point A

⇒ AP = AS …tangents from point to a circle are equal…(i)

BP and BQ are tangents from point B

⇒ BP = BQ …tangents from point to a circle are equal…(ii)

CQ and CR are tangents from point C

⇒ CR = CQ …tangents from point to a circle are equal…(iii)

DR and DS are tangents from point D

⇒ DR = DS …tangents from point to a circle are equal…(iv)

Add equation (i) + (ii) + (iii) + (iv)

⇒ AP + BP + CR + DR = AS + DS + BQ + CQ

From figure AP + BP = AB, CR + DR = CD, AS + DS = AD and BQ + CQ = BC

⇒ AB + CD = AD + BC

Using (a) and (b)

⇒ AB + AB = AD + AD

⇒ 2AB = 2AD

⇒ AB = AD

AB and AD are adjacent sides of parallelogram which are equal hence parallelogram ABCD is a rhombus

Hence if all sides of a parallelogram touch a circle then that parallelogram is a rhombus

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