Conic chords projected by point $P$ form a quadrilateral whose vertex pairs are collinear with $P$ 
Starting with a conic $c$, a point $P$ not on $c$, and points $D,E,F,G$ on $c$, let $D',E',F',G'$ be the second points of intersection of the lines $PD,PE,PF,PG$ with $c$.
Let
$$
\begin{align}
L &= DE\cdot FG \\
M &= D'E'\cdot F'G' \\
N &= FG\cdot F'G' \\
O &= DE\cdot D'E' \\
Q &= DE\cdot F'G' \\
R &= D'E'\cdot FG 
\end{align}
$$
It's easy to show that $O,N$ are on the polar of $P$ wrt $c$.  But here's the question:

show that the triples $P,Q,R$ and $P,L,M$ are respectively collinear.

It's possible to find a projective transformation that takes $P$ to a point at infinity and $c$ to a circle.  Then the lines $DD',EE',\dots$ are parallel and it's easy to show the proposition.  But I'd like a proof that doesn't use that trick.
The question arises from a simplification of 3D configuration, where $c$ is a sphere (or quadric) and two cones with apex $P$ cut $c$ in four planes.  These planes intersect in lines analogous to $L,M,\dots,R$.  I'm hoping a projective proof in 2D will 'lift' to a proof in 3D.
 A: A perspectivity is a projective transformation fixing the points of a line $p$, called the axis of perspectivity, and leaving invariant all the lines through a point P, called the center of perspectivity.
This definition is adapted from Pamfilos.  For more detail and background see Hatton, Projective Geometry, Chapter IV, which uses the terms plane perspective and homology.
Given a point $P$, a line $p$, and points $X,X'$ such that $P,X,X'$ are collinear, there is a unique perspectivity $\mathbf P$ that maps $X$ to $X'$.  For a general point $Y$, write $\mathbf P(Y)$ as $Y'$ (similarly for line $\ell$ and its image $\ell'$).  Then, according to the definition,
$$
\begin{align}
&\bullet\;\;  \text{$P$ is on line $YY'$} \tag{$\star$} \\ 
&\bullet\;\; \text{$\ell,\ell'$ meet on $p$.} \tag{$\star\star$}
\end{align}
$$
Turning to the question, let $P$ be the given point, and $p$ the polar of $P$ with respect to $c$.  Then it is easy to show (using the geometry of polars) that the perspectivity $\mathbf P$ defined by $P,p,D,D'$ maps $c$ to itself, and in particular $E\to E',F->F',G->G'$.
The same is true for lines: $DE\to D'E'$. By ($\star\star$), $DE,D'E'$ meet on $p$.  In other words, $O=DE\cdot D'E'$ is on $p$ and similarly $N$ is on $p$.
$\mathbf P$ takes $L=DE\cdot FG$ to $M=L'=D'E'\cdot F'G'$. By ($\star$), $P$ is on line $LM$.  Similarly $P$ is on line $QR$, and we are done.
This is the conceptual proof I was looking for, and extends nicely to 3D.
