Question

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

Answer 1

Since ABCD is a parallelogram,

AB = CD ......1

BC = AD ......2

So,

DR = DS (Tangents on the circle from point D)

CR = CQ (Tangents on the circle from point C)

BP = BQ (Tangents on the circle from point B)

AP = AS (Tangents on the circle from point A)

Thus,

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

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

CD + AB = AD + BC

Putting the values of equations (1) and (2) in this equation,

2AB = 2BC

AB = BC .......3

Comparing equations (1), (2), and (3),

AB = BC = CD = DA

Thus ABCD is a rhombus.

Answer 2

Answer:

answer in the attachment

hope it helps