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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.

Answer :

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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