Prove that a parallelogram circumscribing a circle is a rhombus.

To prove that a parallelogram circumscribing a circle is a rhombus, we need to delve into some properties of both shapes. A parallelogram is defined as a quadrilateral with opposite sides that are equal and parallel. A circle can be inscribed within a polygon if the polygon is tangential, meaning that it has an incircle that touches each side at exactly one point. For a parallelogram to circumscribe a circle, it must satisfy certain conditions that lead us to conclude that it is indeed a rhombus.

Key Properties of Parallelograms and Circles

First, let's recall some essential properties:

• A parallelogram has opposite sides that are equal in length.
• A circle can be inscribed in a polygon if the sum of the lengths of opposite sides is equal.
• A rhombus is a special type of parallelogram where all four sides are of equal length.

Understanding the Conditions for a Circle to be Inscribed

For a parallelogram to circumscribe a circle, the lengths of the opposite sides must be equal. This means that if we denote the sides of the parallelogram as \(AB\), \(BC\), \(CD\), and \(DA\), we have:

• AB = CD
• BC = DA

Now, since the circle is inscribed, we can express the relationship of the sides in terms of their lengths. Let’s denote the lengths of the sides as follows:

• Let \(AB = a\)
• Let \(BC = b\)

Since the opposite sides are equal, we have \(CD = a\) and \(DA = b\). The condition for the circle to be inscribed is that the sum of the lengths of opposite sides must be equal:

Thus, we have:

a + b = a + b

Deriving the Rhombus Condition

Now, let’s analyze what happens if we assume that the parallelogram is tangential. The tangential condition implies that the lengths of the sides must be equal:

• If \(AB = CD\) and \(BC = DA\), then for the circle to be inscribed, we must have \(a + b = a + b\) satisfied for all sides.

However, since the circle is inscribed, we can also derive that:

• AB + CD = BC + DA

Substituting the values, we get:

a + a = b + b

This simplifies to:

2a = 2b

From this equation, we can conclude that:

a = b

Conclusion: The Parallelogram is a Rhombus

Since we have established that all sides of the parallelogram are equal (i.e., \(AB = BC = CD = DA\)), we can conclude that the parallelogram is indeed a rhombus. Therefore, any parallelogram that circumscribes a circle must have all sides of equal length, confirming that it is a rhombus.

This proof elegantly ties together the properties of tangential quadrilaterals and the specific characteristics of parallelograms, leading us to a clear conclusion about the nature of the shape in question.