Question

Prove that the parallelogram circumscribing a circle is a rhombus.

Hint:

A parallelogram that circumscribes a circle has equal sides, and therefore it is a rhombus.

Answer 1

Step 1: Consider the parallelogram.}
Let \( ABCD \) be a parallelogram that circumscribes a circle. This means the incircle of the parallelogram touches all its sides.
Step 2: Property of tangents.}
Let the points of contact of the incircle with the sides \( AB, BC, CD, \) and \( DA \) be \( P, Q, R, \) and \( S \) respectively. From the tangent-secant property, we know that the tangents drawn from an external point to a circle are equal in length. Hence: \[ AP = AS, \quad BP = BQ, \quad CQ = CR, \quad DR = DS \]
Step 3: Use the sum of tangents.}
Now, the sum of the lengths of the opposite sides of a parallelogram are equal. So, we have: \[ AB + CD = AD + BC \]
Step 4: Show that the sides are equal.}
From the tangency conditions, we know that: \[ AB = AP + BP = AS + AR = AD \] Similarly: \[ BC = BQ + CQ = BR + CR \] Therefore, \( AB = BC = CD = DA \), showing that all four sides of the parallelogram are equal in length.

Step 5: Conclusion.}
Since all the sides of the parallelogram are equal, it must be a rhombus. % Final Answer Final Answer:
Thus, the parallelogram circumscribing a circle is a rhombus.