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⇒ In the given figure , the following things should be noted -

                                        ⇒ ABCD is a rhombus

                                        ⇒ Diagonals of the rhombus are AC and BD
                                           which intersect at O.


''The circle drawn with any side of a rhombus as diameter''


⇒ ∠ AOB = ∠ BOC = ∠ COD = ∠ AOD = 90 [ Diagonal of the rhombus bisect                                                                      each other at  90°]

Now in the given four circles , we have four diameters -  AB , BC , CD , DA.

⇒ All this diameters passes through O. { Angles in a semi circle is right                                                                     angle }

Therefore by the above information , we proved that the circles described in the four sides of a rhombus as a diameter, pass through the point of intersection of its diagonal.
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