Prove that a cyclic parallelogram is a rectangle.

Hint: To solve this question we have to prove that an angle of parallelogram measures $ 90{}^\circ $ because for a parallelogram to be a rectangle one of its angle measures $ 90{}^\circ $ . We will use the fact that the sum of opposite angles of a cyclic parallelogram equal to $ 180{}^\circ $.

Complete step by step answer:
We have been given a cyclic parallelogram.
We have to prove that it is a rectangle.

We have given a cyclic parallelogram $ ABCD $ .
Now, we know that the sum of opposite angles of a cyclic parallelogram is equal to $ 180{}^\circ $.
So we have $ \angle A+\angle C=180{}^\circ $
Now, we know that opposite angles of a parallelogram are equal in measure, so we get
 $ \Rightarrow \angle A=\angle C $
Substituting the values we get
 $ \Rightarrow \angle A+\angle A=180{}^\circ $
Solving further we get
 $ \begin{align}
  & \Rightarrow 2\angle A=180{}^\circ \\
 & \Rightarrow \angle A=\frac{180{}^\circ }{2} \\
 & \Rightarrow \angle A=90{}^\circ \\
\end{align} $
Now, we know that if one of the angle of a cyclic parallelogram is equal to $ 90{}^\circ $ , then we can say that it is a rectangle.
Hence proved that a cyclic parallelogram is a rectangle.
Note:
An alternate way to solve this question is as follows-
We can say from the given diagram that DC is the diameter of the given circle and we know that the diameter of a circle subtends the $ 90{}^\circ $ angle at the circumference of the circle. By using this fact we can prove that an interior angle of the parallelogram is $ 90{}^\circ $, hence it is a rectangle.