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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.

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. $\square$