Question

prove that all four circles drawn with the sides of a rhombus as diameters pass to a common point

Answer 1

mark points abcd of rhombus. then take radius r/2 then write eqn of 4 circles and then using elimination method find the point. this proves your question

Answer 2

Let ABCD be the rhombus with AC and BD as diagonals intersecting at point O.
Recall that, diagonals of a rhombus bisect each other at right angles.
∴ ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°
Consider the circle with AB as diameter passes through O.  [Since angle in a semi-circle is a right angle]
Similarly we can prove for other sides BC, CD and DA also.
Therefore, the circles with four sides of a rhombus as diameter, pass through the point of intersection of its diagonals