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$A,B,C$ are three collinear points and $P$ is a point not in the line $AB$: $AFE,BFD$ and $CED$ are perpendiculars to $PA,PB,PC$ respectively. Prove that $P,D,E,F$ are concyclic.

I attempted to do this using angle chasing. But I can't really understand why the lines are being defined by 3 points, since in my diagrams, this is simply not happening. So I need some help with this.

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  • $\begingroup$ Could you tell us where $D$,$E$ and $F$ are? Your second sentence seems a little unclear $\endgroup$ – Benson Lin Oct 10 '16 at 12:59
  • $\begingroup$ I am having some difficulty with that as well. $\endgroup$ – SANTANU NANDI Oct 10 '16 at 13:25
  • $\begingroup$ $F$ is the intersection point of the perpendicular to $PA$ at $A$ and the perpendicular to $PB$ at $B$, etc. $\endgroup$ – Lozenges Oct 10 '16 at 14:28
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Simson line. Line $AB \equiv AC$ is the Simson line for triangle $EFD$ from the point $P$. By the theorem for the Simson line, the point $P$ lies on the circumcircle of $EFD$ if and only if its three orthogonal projections onto the extended edges of the triangle $EFD$ are collinear, which is the case here.

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