Question

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

Answer 1

see explanation.

Explanation:

As shown in the diagram, `ABCD` is a parallelogram, `WXYZ` are points of tangency.
As `ABCD` is a parallelogram,
`=> AB=CD, and AD=BC`
As two tangent segments to a circle from an external point are equal,
`=> AW=AZ, BW=BX, CY=CX, DY=DZ`
`=> AW+BW+CY+DY=AZ+BX+CX+DZ`
`=> AB+CD=AD+BC`
`=> 2AB=2AD`
`=> AB=AD`
Since `AB and AD` are adjacent sides of a parallelogram,
`=>` parallelogram `ABCD` is a rhombus.