Q) Prove that the parallelogram circumscribing a circle is a rhombus.

Ans: 

Step 1: Let’s start by making a diagram for the given question:

Here ABCD is a parallelogram and its is circumscribing a circle with center O.

This parallelogram is touching the circle at points P, Q, R and S.

Step 2: Let’s connect O with P and Q.

By tangents property, we know that the tangents drawn on a circle from an external point are always equal,

∴ from Point A: AP = AS ………….. (i)

Similarly, from Point B: BP = BQ …………  (ii)

and from Point C: CR = CQ …………  (iii)

and from Point D: DR = DS …………  (iv)

Step 3: Let’s add all the 4 equations, we get:

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

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

∴ AB + CD = AD  + BC ……. (v)

Step 4: Its is given that ABCD is a parallelogram,

∴ AB = CD and AD = BC  (∵ opposite sides of a parallelogram are always equal)

Therefore, equation (v) now becomes,

∴ AB + CD = AD  + BC

∴ AB + AB = BC  + BC

∴ 2 AB = 2 BC

∴ AB = BC

∴ AB = BC = CD = AD

Therefore, parallelogram ABCD is a Rhombus.

Hence Proved !

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