## The separation theorem

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Anindya Biswas
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The separation theorem
Let \$A,B, C, D\$ be \$4\$ distinct points on the plane. Every circle going through \$A,C\$ intersects with every circle going through \$B,D\$. Prove that \$A,B, C, D\$ are either concyclic or collinear.
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Mehrab4226
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### Re: The separation theorem

Anindya Biswas wrote:
Fri Aug 13, 2021 10:52 am
Let \$A,B, C, D\$ be \$4\$ distinct points on the plane. Every circle going through \$A,C\$ intersects with every circle going through \$B,D\$. Prove that \$A,B, C, D\$ are either concyclic or collinear.
Let us assume that there is 4 points that are neither on a line nor on a circle, but has that property.

Let the points are \$A,B,C,D\$ as above. Clearly the perpendicular bisector of \$A,C\$ and \$B,D\$ are not the same as \$A,B,C,D\$ are not colinear.

If they intersect, we can take the intersection of the perpendicular bisectors as the centre and draw 2 concentric circles not having the property. The two circles are different because they cannot be the same by our assumption.

If they are parallel, the lines \$AC\$ and \$BD\$ are parallel as well. We can take the points \$X,Y\$ on the perpendicular bisector on the mid region of the two parallel lines \$AC\$ and \$BD\$. Circles \$ACX\$ and \$BDY\$ have no common point. A contradiction. \$\square\$
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