Question

Prove that If the sides of a parallelogram touch a circle, then the parallelogram is a rhombus.

Answer 1

Given - A circle touches the sides AB, BC, CD and DA of ||gm. ABCD touches the circle P,Q, R, and S respectively.

To Prove - ABCD is a rhombus

Proof - Since A, AP and AS are the tangents to the circle
$\therefore$ AP=AS ...(i)

Similarly, we can prove that,
BP = BQ ...(ii)
CR=CQ ...(iii)
and DR = DS ...(iv)

Adding, (i) to (iv )we get:
AP + BD + CR + DR = AS + BQ + CQ + DS
$\Rightarrow$ AP + PB + CR + R = AS + SD + BQ + QC
$\Rightarrow$ AB + CD = AD + BC

But AB = CE and BC = AD ...(iv)
(opposite sides of a ||gm)
$\therefore$ AB+AB=BC+BC
$\Rightarrow$ 2AB = 2BC
$\Rightarrow$ AB=BC ...(v)

From (v) and (vi)
AB = BC = CD = DA
Hence ABCD is a rhombus.