Question

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

Answer 1

$\bf \: SOLUTION : -$

Let ABCD be a parallelogram such that its side touch a circle with centre O.

We know that the tangents to a circle from an exterior point are equal in length.

Therefore, AP = AS [From A] ...(i)

BP = BQ [From B] ...(ii)

CR = CQ [From C] ...(iii)

And DR = DS [From D] ...(iv)

Adding (i), (ii), (iii) and (iv), we get,

AP + BP + CR + DR = AS + BQ + CQ + DS

→ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

→ AB + CD = AD + BC

→ 2 AB = 2 BC

→ AB = BC

Therefore, AB = BC = CD = AD

Thus, ABCD is a rhombus.