Prove that the parallelogram circumscribing a circle is a rhombus

Step 1: Draw a imaginary diagram according to the question
NOTE: Since ABCD is a parallelogram,
AB = CD ..............(1)
BC = AD ................(2)
Step 2: Mark the circle touching points with the sides of the parallelogram.
(S, P, Q, and R tangencies )
Step 3: Observe the figure and Write the equal lengths
NOTE: Lengths of tangents from an external points are equal.
EXAMPLE: 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)
Step 4: Adding all these equations, (Obtained in step 3)
EXAMPLE: DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
Step 5: Putting the values of equations (1) and (2) in the equation obtained in step 3, (See the step 1 for Eq (1)&(2))
EXAMPLE: 2AB = 2BC
AB = BC …......(3)
Step 6: Compare equation 1, 2 and 3 (See step 1 and step 5)
NOTE: All the sides of the parallelogram are equal so, it is a rhombus
EXAMPLE: AB = BC = CD = DA
Hence, ABCD is a rhombus.