Question

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Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Let ABCD is a parallelogram circumscribing with a circle.    

Now since ABCD is a parallelogram.

So AB = CD and AD = BC

Again, AP = AS .........1  

Similarly

PB =BQ .......2

CR = CQ ......3

DR  = DS......4

Adding all 4 equations, we get

=> AP + PB + DR + CR = AS + DS + BQ + CQ

=> (AP + PB) + (DR + CR) = (AS + DS) + (BQ + CQ)

=> AB + CD = AD + BC

=> AB +AB = AD + AD   (since AD = BC and AB = CD)

=> 2AB = 2AD

=> AB = AD

Since AB and AD are adjacent sides of parallelogram.

Hence, parallelogram ABCD is a rhombus.

Hope it helps!