We know that Five points determine a conic and Two Points Determine a Line. Here I found a simple construct of a conic through $7$ points (in PS I note that how the conic through thirteen points) and a line through $4$ points as follows:
Let $ABC$ be a triangle and $P$ be arbitrary point in the plane, let $A'$, $B'$, $C'$ lie on the line through $P$ and parallel to $BC$, $CA$, $AB$ respectively such that $AA' \parallel BB' \parallel CC'$. I am looking for a proof that
- Seven points $A$, $B$, $C$, $A'$, $B'$, $C'$ and $P$ lie on a conic
- Four points: midpoints of $AA'$, $BB'$, $CC'$ and center of the conic are collinear?
PS: Three lines: The line through midpoints of $BC$ and $PA'$, The line through midpoints of $CA$ and $PB'$ and The line through midpoints of $AB$ and $PC'$ are concurrent at the center of the conic (denote the point of concurrence is $O$). Then easily show that reflection of $A, B, C, A', B', C'$ in $O$ also lie on the conic. This mean the conic through $13$ points.