Prove that the circle described with the four sides of a rhombus as diameters pass through the point of intersection of its diagonals.

Consider $ABCD$ as a rhombus

We know that the diagonals $AC$ and $BD$ intersect at the point $O$

From the figure we know that the diagonals of the rhombus bisect at right angles 

It can be written as

Thus, $\angle BOC$ lies in the circle

We know that the circle can be drawn with $BC$ as the diameter having the centre $O$

In the same way, all the circles with $AB,AD$ and $CD$ as diameters will pass through the centre $O$.

Therefore , it is proved that the circles described with the four sides of a rhombus as diameters pass through the point of intersection of its diagonals.