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# Prove that all four circles drawn with the sides of a rhombus as diameters pass through a common point.Prove that this is true for any quadrilateral with adjacent sides equal, as in the picture.

Let ABCD be the rhombus with AC and BD as diagonals intersecting at point O

We know 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 Consider the circle with BC as diameter passes through O. [Since angle in a semi-circle is a right angle]

Similarly Consider the circle with DC as diameter passes through O. [Since angle in a semi-circle is a right angle]

Similarly Consider the circle with AD as diameter passes through O. [Since angle in a semi-circle is a right angle]

Therefore, the circles with four sides of a rhombus as diameter, pass through the point of intersection of its diagonals.

Next part

First, we will consider the right side of AC

Consider a Δ ACD,

Two circles are drawn with CD and AD as a diameter.

Let they intersect each other at P and let P does not lie on AC

Join DP

DPA = 90° (angle subtended by semi circle)

DPC = 90° (angle subtended by semi circle)

APC = DPA + DPC = 90° + 90° = 180°

Therefore, APC is a straight line and hence our assumption is wrong.

Thus point P lies on third side AC on Δ ACD

Now, we will consider the left side of AC

Consider a Δ ABC

Two circles are drawn with CB and AB as a diameter.

Let they intersect each other at P and let P does not lie on AC

Join BP

BPA = 90° (angle subtented by semi circle)

BPC = 90° (angle subtended by semi circle)

APC = BPA + BPC = 90° + 90° = 180°

Therefore, APC is a straight line and hence our assumption is wrong.

Thus, point P lies on third side AC on Δ ABC

Therefore, the circles with four sides of a quadrilateral as diameter, pass through the point of intersection of its diagonals.

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