Q. 85.0( 1 Vote )

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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