Prove that the circle drawn with any side of a rhombus as diameter passes

Solution:

Let ABCD be a rhombus in which diagonals are intersecting at point O and a circle is drawn while taking side CD as its diameter. We know that a diameter subtends 90° on the arc.

$\therefore \angle C O D=90^{\circ}$

Also, in rhombus, the diagonals intersect each other at $90^{\circ}$.

$\angle A O B=\angle B O C=\angle C O D=\angle D O A=90^{\circ}$

Clearly, point $O$ has to lie on the circle.