Prove that the parallelogram circumscribing a circle is a rhombus.

Solution: Please refer to video for the figure.
For parallelogram $ABCD$
$AB=CD\text{and}AD=BC$
We know, tangents drawn from a point to a circle are always equal.So,
$AE=AH,EB=BF,GC=CF,DG=HD$
$AE+EB+CG+DG=AH+HD+CF+BF$
$AB+CD=AD+BC$
$2AB=2BC\Rightarrow AB=BC$
Similarly, $AD=CD$
Thus, all sides are equal.So, ABCD is a rhombus.