Prove that a cyclic parallelogram is a rectangle.

Let $ABCD$ be a parallelogram such that its diagonals $AC$ and $BD$ are the diameters of the circle through the vertices $A, B, C$ and $D$.

We know that the angle in a semi-circle is a right angle.

Therefore,

$\angle ADC = 90^o$ and $\angle ABC = 90^o$ 

$\angle BCD = 90^o$ and $\angle BAD = 90^o$

We know that a parallelogram with one right angle is a rectangle. 

Therefore, $ABCD$ is a rectangle. 

Hence proved.