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

Let the rhombus be ABCD, with AB as one side. Let the diagonals AC and BD intersect at point O. We want to prove that the circle with AB as diameter passes through O.

In a rhombus, the diagonals are perpendicular bisectors of each other. Therefore, ∠AOB = 90°.

Consider the triangle AOB. Since the diagonals bisect each other, AO = OC and BO = OD. Also, in a rhombus, all sides are equal in length, so AB = BC = CD = DA.

In right-angled triangle AOB, if we draw a circle with AB as the diameter, then by the property of a circle, any angle subtended by the diameter on the circumference is a right angle. Therefore, the point O, which subtends a right angle ∠AOB at the circumference, must lie on the circle with diameter AB.

Similarly, we can show that if we draw a circle with any other side (BC, CD, or DA) as the diameter, the point O will also lie on that circle because the diagonals are perpendicular bisectors, and the angle subtended by the diameter at the intersection point will always be 90°.