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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.

 Video Solution
Circles
Ex 10.2 | Question 11

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

\[\begin{align}  & BP+CR+DR+AP \\  & =BQ+CQ+DS+AS \\ \end{align}\]
On re-grouping,

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

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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