Proof that a Parallelogram Circumscribing a Circle is a Rhombus

Introduction

A parallelogram is a two-dimensional quadrilateral with opposite sides parallel. A circle is a two-dimensional shape with all points equidistant from a fixed point called the center. A parallelogram that circumscribes a circle means that the parallelogram has four sides tangent to the circle.

In this article, we will prove that a parallelogram that circumscribes a circle is a rhombus. A rhombus is a parallelogram with all four sides equal in length.

Proof

To prove that a parallelogram that circumscribes a circle is a rhombus, we need to show that all four sides of the parallelogram are equal in length.

Let ABCD be a parallelogram that circumscribes a circle with center O. Let the points of tangency of the parallelogram with the circle be E, F, G, and H.

Since ABCD circumscribes the circle, we have:

• AE = AF (by definition of tangency)
• BE = BF (by definition of tangency)
• CG = CH (by definition of tangency)
• DG = DH (by definition of tangency)

Since AE = AF and BE = BF, we have AB = 2AE. Similarly, since CG = CH and DG = DH, we have CD = 2CG.

Since AB = 2AE and CD = 2CG, we have AB = CD. Similarly, since AD = 2AF and BC = 2BF, we have AD = BC.

Therefore, all four sides of the parallelogram ABCD are equal in length, which means that ABCD is a rhombus.

Q.E.D.

Corollary

As a corollary to the above theorem, we have the following:

• The diagonals of a parallelogram that circumscribes a circle are perpendicular to each other.

Proof

Let ABCD be a parallelogram that circumscribes a circle with center O. Let the diagonals of the parallelogram be AC and BD.

Since ABCD circumscribes the circle, we have:

• AE = AF (by definition of tangency)
• BE = BF (by definition of tangency)
• CG = CH (by definition of tangency)
• DG = DH (by definition of tangency)

Since AE = AF and BE = BF, we have AB = 2AE. Similarly, since CG = CH and DG = DH, we have CD = 2CG.

Since AB = 2AE and CD = 2CG, we have AB = CD. Similarly, since AD = 2AF and BC = 2BF, we have AD = BC.

Therefore, all four sides of the parallelogram ABCD are equal in length, which means that ABCD is a rhombus.

Since ABCD is a rhombus, the diagonals AC and BD are perpendicular to each other.

Q.E.D.

Applications

The fact that a parallelogram that circumscribes a circle is a rhombus has many applications in geometry. For example, it can be used to prove that the diagonals of a rhombus are equal in length. It can also be used to find the area of a rhombus.

Conclusion

In this article, we have proven that a parallelogram that circumscribes a circle is a rhombus. This result has many applications in geometry.

Additional Resources

• Parallelograms and Rhombuses
• Circles and Tangents
• The Pythagorean Theorem

Keywords

• Parallelogram
• Rhombus
• Circle
• Tangent
• Diagonals
• Geometry