How i prove the parallelogram of question no.11 of ncert class 10th book of exercise 10.2, please give me the simple way for to prove the parallelogram Prove that the parallelogram circumscribing a circle is a rhombus.

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Consider a parallelogram ABCD which is circumscribing a circle with a center O. Now, since ABCD is a parallelogram, AB = CD and BC = AD.

From the above figure, it is seen that,

(i) DR = DS

(ii) BP = BQ

(iii) CR = CQ

(iv) AP = AS

These are the tangents to the circle at D, B, C, and A respectively.

Adding all these we get,

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

By rearranging them we get,

(BP+AP)+(DR+CR) = (CQ+BQ)+(DS+AS)

Again by rearranging them we get,

AB+CD = BC+AD

Now, since AB = CD and BC = AD, the above equation becomes

2AB = 2BC

∴ AB = BC

Since AB = BC = CD = DA, it can be said that ABCD is a rhombus.