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

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## 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.

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