A parallelogram has an inscribed circle touching all its four sides. How do you prove that it is a rhombus?

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A parallelogram has an inscribed circle touching all its four sides. How do you prove that it is a rhombus?

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Answer 1 · 2017-02-18T14:51:55

see explanation.

Explanation:

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As shown in the diagram, $A B C D$ $ABCD$ is a parallelogram, $W X Y Z$ $WXYZ$ are points of tangency.
As $A B C D$ $ABCD$ is a parallelogram,
$\Rightarrow A B = C D, and A D = B C$ $=> AB=CD, and AD=BC$
As two tangent segments to a circle from an external point are equal,
$\Rightarrow A W = A Z, B W = B X, C Y = C X, D Y = D Z$ $=> AW=AZ, BW=BX, CY=CX, DY=DZ$
$\Rightarrow A W + B W + C Y + D Y = A Z + B X + C X + D Z$ $=> AW+BW+CY+DY=AZ+BX+CX+DZ$
$\Rightarrow A B + C D = A D + B C$ $=> AB+CD=AD+BC$
$\Rightarrow 2 A B = 2 A D$ $=> 2AB=2AD$
$\Rightarrow A B = A D$ $=> AB=AD$
Since $A B and A D$ $AB and AD$ are adjacent sides of a parallelogram,
$\Rightarrow$ $=>$ parallelogram $A B C D$ $ABCD$ is a rhombus.

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A parallelogram has an inscribed circle touching all its four sides. How do you prove that it is a rhombus?

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Explanation:

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