# Prove that the parallelogram circumscribing a circle is a rhombus.

Last updated date: 24th Mar 2023

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Hint- Here, we will be using the concept of Two-Tangent Theorem.

Let us draw a parallelogram ABCD with sides AB equal and parallel to side CD and side BC equal and parallel to side DA. Also, draw a circle circumscribed by parallelogram ABCD as shown in figure.

As already discussed, ${\text{AB}} = {\text{CD }} \to {\text{(1)}}$ and ${\text{BC}} = {\text{AD }} \to {\text{(2)}}$

From the figure, it is clear that all the sides of the parallelogram are tangent to the given circle.

As we know that according to the Two-Tangent Theorem, if two tangent segments are drawn to one circle from the same external point then the length of both the tangent segments will be equal.

From the figure, we can say that two tangent segments AP and AS are drawn from the same external point A to the given circle. Similarly, two tangent segments BP and BQ are drawn from the same external point B to the given circle. Similarly, two tangent segments CQ and CR are drawn from the same external point C to the given circle. Similarly, two tangent segments DS and DR are drawn from the same external point D to the given circle

.

Therefore, ${\text{AP}} = {\text{AS}}$, ${\text{BP}} = {\text{BQ}}$, ${\text{CR}} = {\text{CQ}}$ and ${\text{DR}} = {\text{DS}}$.

Now adding all the above four equations, we get

${\text{AP}} + {\text{BP}} + {\text{CR}} + {\text{DR}} = {\text{AS}} + {\text{BQ}} + {\text{CQ}} + {\text{DS }} \to {\text{(3)}}$

From the figure, we can write

${\text{AP}} + {\text{BP}} = {\text{AB}}$, ${\text{BQ}} + {\text{CQ}} = {\text{BC}}$, ${\text{CR}} + {\text{DR}} = {\text{CD}}$ and ${\text{DS}} + {\text{AS}} = {\text{AD}}$

Using the above equations in equation (3), we get

\[\left( {{\text{AP}} + {\text{BP}}} \right) + \left( {{\text{CR}} + {\text{DR}}} \right) = \left( {{\text{AS}} + {\text{DS}}} \right) + \left( {{\text{BQ}} + {\text{CQ}}} \right) \Rightarrow {\text{AB}} + {\text{CD}} = {\text{AD}} + {\text{BC }} \to {\text{(4)}}\]

Now using equations (1) and (2) in equation (4), we get

\[ \Rightarrow 2{\text{AB}} = 2{\text{BC}} \Rightarrow {\text{AB}} = {\text{BC}}\]

So, we can say that \[{\text{AB}} = {\text{BC}} = {\text{CD}} = {\text{AD}}\]

We know that a rhombus is a quadrilateral whose four sides all have the same length.Since, the above condition proves that parallelogram ABCD which is circumscribing the circle is a rhombus (i.e., ABCD is a rhombus).

Note- In this particular problem, in order to prove that the parallelogram is a rhombus we have to prove that all the sides of the parallelogram circumscribing the circle will be equal.

Let us draw a parallelogram ABCD with sides AB equal and parallel to side CD and side BC equal and parallel to side DA. Also, draw a circle circumscribed by parallelogram ABCD as shown in figure.

As already discussed, ${\text{AB}} = {\text{CD }} \to {\text{(1)}}$ and ${\text{BC}} = {\text{AD }} \to {\text{(2)}}$

From the figure, it is clear that all the sides of the parallelogram are tangent to the given circle.

As we know that according to the Two-Tangent Theorem, if two tangent segments are drawn to one circle from the same external point then the length of both the tangent segments will be equal.

From the figure, we can say that two tangent segments AP and AS are drawn from the same external point A to the given circle. Similarly, two tangent segments BP and BQ are drawn from the same external point B to the given circle. Similarly, two tangent segments CQ and CR are drawn from the same external point C to the given circle. Similarly, two tangent segments DS and DR are drawn from the same external point D to the given circle

.

Therefore, ${\text{AP}} = {\text{AS}}$, ${\text{BP}} = {\text{BQ}}$, ${\text{CR}} = {\text{CQ}}$ and ${\text{DR}} = {\text{DS}}$.

Now adding all the above four equations, we get

${\text{AP}} + {\text{BP}} + {\text{CR}} + {\text{DR}} = {\text{AS}} + {\text{BQ}} + {\text{CQ}} + {\text{DS }} \to {\text{(3)}}$

From the figure, we can write

${\text{AP}} + {\text{BP}} = {\text{AB}}$, ${\text{BQ}} + {\text{CQ}} = {\text{BC}}$, ${\text{CR}} + {\text{DR}} = {\text{CD}}$ and ${\text{DS}} + {\text{AS}} = {\text{AD}}$

Using the above equations in equation (3), we get

\[\left( {{\text{AP}} + {\text{BP}}} \right) + \left( {{\text{CR}} + {\text{DR}}} \right) = \left( {{\text{AS}} + {\text{DS}}} \right) + \left( {{\text{BQ}} + {\text{CQ}}} \right) \Rightarrow {\text{AB}} + {\text{CD}} = {\text{AD}} + {\text{BC }} \to {\text{(4)}}\]

Now using equations (1) and (2) in equation (4), we get

\[ \Rightarrow 2{\text{AB}} = 2{\text{BC}} \Rightarrow {\text{AB}} = {\text{BC}}\]

So, we can say that \[{\text{AB}} = {\text{BC}} = {\text{CD}} = {\text{AD}}\]

We know that a rhombus is a quadrilateral whose four sides all have the same length.Since, the above condition proves that parallelogram ABCD which is circumscribing the circle is a rhombus (i.e., ABCD is a rhombus).

Note- In this particular problem, in order to prove that the parallelogram is a rhombus we have to prove that all the sides of the parallelogram circumscribing the circle will be equal.

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