Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

(N/A) Let $ABCD$ be a rhombus whose diagonals $AC$ and $BD$ intersect at $O$. We know that the diagonals of a rhombus bisect each other at right angles, so $\angle AOB = 90^\circ$.
Consider a circle drawn with side $AB$ as the diameter. Let $Q$ be the midpoint of $AB$. Then $QA = QB = \text{radius}$.
In $\triangle AOB$, $\angle AOB = 90^\circ$. Since $Q$ is the midpoint of the hypotenuse $AB$ of the right-angled triangle $\triangle AOB$, the distance from the midpoint of the hypotenuse to the vertices is equal to half the length of the hypotenuse.
Therefore, $QO = QA = QB$.
Since the distance of point $O$ from the center $Q$ is equal to the radius ($QA$ or $QB$), the point $O$ must lie on the circle drawn with $AB$ as the diameter.
Thus, the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.