Prove that the parallelogram circumscribing a circle is a rhombus.

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Step 1: Parallelogram ABCD circumscribing a circle with tangent points P, Q, R, S
Step 2: $AP = AS$, $BP = BQ$, $CR = CQ$, $DR = DS$ (tangent lengths from external point)
Step 3: Adding: $AP + BP + CR + DR = AS + BQ + CQ + DS \Rightarrow AB + CD = AD + BC$
Step 4: But $ABCD$ is a parallelogram $\Rightarrow AB = CD$ and $AD = BC$. $	herefore 2AB = 2AD \Rightarrow AB = AD$. Hence $ABCD$ is a rhombus.

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