# Ex.10.2 Q11 Circles Solution - NCERT Maths Class 10

## Question

Prove that the parallelogram circumscribing a circle is a rhombus.

## Text Solution

**To prove:**

The parallelogram circumscribing a circle is a rhombus

**Reasoning:**

\({ABCD}\) is a parallelogram. Therefore, opposite sides are equal.

\[\begin{align}\therefore {AB} & = {CD} \\ {BC} & = {AD} \end{align}\]

According to **Theorem 10.2 :** The lengths of tangents drawn from an external point to a circle are equal.

**Steps :**

Therefore,

\(BP = BQ\) (Tangents from point \(B\))…… (1)

\(CR = CQ\) (Tangents from point \(C\))…… (2)

\(DR = DS\) (Tangents from point \(D\))…… (3)

\(AP = AS\) (Tangents from point \(A\))……. (4)

Adding \((1) + (2) + (3) + (4)\)

\[{BP + CR + DR + AP = BQ + CQ + DS + AS}\]

On re-grouping,

\[\begin{align} BP + AP + CR + DR &= BQ + CQ + DS + AS\\AB + CD &= BC + AD \end{align}\]

Substitute \({CD = AB}\) and \({AD = BC}\) since \({ABCD}\) is a parallelogram, then

\[\begin{align}{ A B + A B} & = {B C + B C} \\ {2 A B} & ={ 2 B C} \\ {A B} & = {BC} \end{align}\]

\(\therefore {AB=BC=CD=DA}\)

This implies that all the four sides are equal.

Therefore, the parallelogram circumscribing a circle is a rhombus.