Question

Prove that a cyclic parallelogram is a rectangle.

Answer 1

We are given that a parallelogram is cyclic, which means all its vertices lie on the circumference of a circle. We need to prove that a cyclic parallelogram is a rectangle.

Step 1: Definition of a Cyclic Parallelogram

A cyclic parallelogram is a quadrilateral whose four vertices lie on a common circle. This implies that all the angles of the cyclic parallelogram are subtended by the same circle.

Step 2: Properties of Parallelograms

In a parallelogram, opposite angles are equal. Let the angles of the parallelogram be denoted as: \[ \angle ABC = \angle CDA = x \quad \text{and} \quad \angle DAB = \angle BCD = y \]

Step 3: Sum of Angles in a Cyclic Quadrilateral

Since the parallelogram is cyclic, the sum of opposite angles of a cyclic quadrilateral is \(180^\circ\). Hence, we have: \[ \angle ABC + \angle DAB = 180^\circ \] Substituting the values of the angles: \[ x + y = 180^\circ \]

Step 4: Conclusion - The Parallelogram is a Rectangle

In a rectangle, the angles are all \(90^\circ\). Since we know that the sum of the adjacent angles in a cyclic quadrilateral is \(180^\circ\), we conclude that: \[ x = 90^\circ \quad \text{and} \quad y = 90^\circ \] Therefore, all angles in the cyclic parallelogram are \(90^\circ\), making it a rectangle.

Final Conclusion:

A cyclic parallelogram is a rectangle because the sum of adjacent angles in a cyclic quadrilateral is \(180^\circ\), and in this case, each angle is \(90^\circ\).