Question

If all sides of a parallelogram touch a circle, show that the parallelogram is a rhombus

Answer 1

Let $ABCD$ be a parallelogram.
Parallelogram touches the circle at $P,Q,R,S$
$AP=AS$ .... (1)
$PB=BQ$ .... (2)
$CR=CQ$ .... (3)
$DR=DS$ .... (4)
Now add (1), (2), (3) and (4),
$AP+PB+CR+DR=AS+BQ+CQ+DS$
$\Rightarrow AP+PB+CR+DR=AS+DS+BQ+CQ$
$\Rightarrow AB+CD=AD+BC$
$\Rightarrow AB+AB=BC+BD$ ....$(\because$ In a parallelogram $AB=CD,BC=AD)$
$\Rightarrow 2AB=2BC$
$\Rightarrow AB=BC$
$\therefore$ the parallelogram becomes a rhombus.