**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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