Question

Prove that the parallelogram circumscribing a circle is a rhombus​

Answer 1

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 (since AP and AS are two tangents drawn from external point)

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.

Hense parallelogram ABCD is a rhombus.

Hope it helps u....

Answer 2

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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.

Ncert solutions class 10 chapter 10-13

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.