Prove that the parallelogram circumscribing a circle is a rhombus.

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Step 1: Let $ABCD$ be a parallelogram 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 are equal)
Step 3: Adding: $(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) \Rightarrow AB + CD = AD + BC$
Step 4: Since $AB = CD$ and $AD = BC$ (parallelogram), $2AB = 2BC \Rightarrow AB = BC$. $	herefore ABCD$ is a rhombus.

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