Question

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

Answer 1

Let ABCD be a parallelogram whose sides AB, BC, CD and DA touch a circle at the points P, Q, R and S respectively.

Since the lengths of tangents drawn from an external point to a circle are equal, we have

AP = AS, BP = BQ, CR = CQ and DR = DS.

AB + CD = AP + BP + CR + DR

= AS + BQ + CQ + DS

= (AS + DS) + (BQ+CQ)

= AD + BC

Now, AB + CD = AD + BC

$\Rightarrow$ 2AB = 2BC

$\Rightarrow$ AB = BC

AB = BC = CD = AD.

Hence, ABCD is a rhombus.