One line solution with projective geometry

Question

I am trying to solve the following problem

Let $A,B,C,D$ be four points in the plane.Let lines $AC$ and $BD$ meet at $P$ , lines $AB$ and $CD$ meet at $Q$, and lines $BC$ and $DA$ meet at $R$. Let line through $P$ parallel to $QR$ meet lines $AB$ and $CD$ at $X$ and $Z$. Show that $P$ is the midpoint of $XZ$.

There is a hint from the author that it can be solved in one line using projective geometry and specifically perspectivity at $Q$.

I can't find this one liner does anyone have any ideas?

Answer 1

Let $S=PQ\cap DA$, and let $P_{\infty}=XY\cap QR$. Then $-1=(A,D;S,R)\stackrel{Q}{=}(X,Y;P,P_{\infty})$, so $P$ is the midpoint of $\overline{XY}$.

Answer 2

Consider the projective involution (maybe it is called perspectivity, I don't remember the terminology) with center $P$ and axis $QR$ which fixes line $QR$ pointwise and maps point $A$ to $C$. Then line $AB$ is mapped to $DC$ and $AD$ to $BC$. Consequently $X$ is mapped to $Z$. Since $XZ$ is parallel to the axis $QR$, the center $P$ is the midpoint of $XZ$.

I write a bit more than average (it's just me, I am a bit wordy), but I believe this should be the spirit of the proof.