So far I have built it this way but I'm not sure if it's enough:
Given $\triangle PQR$ we choose two arbitrary sides of this triangle and place a point anywhere in each side. Lets say $D$ in $RQ$ and $E$ in $RP$, then we build the harmonic conjugate to $D$ respect to $R$, $Q$ and the harmonic conjugate to $E$ with respect to $R$, $P$. We name these harmonic conjugates $B$ and $F$ respectively. We will have that $D$, $B$ and $E$, $F$ are two pairs of opposite vertices of our complete quadrilateral so to find the other two vertices we simply trace the lines $BE$, $BF$, $DE$, and $DF$ and see where they come together.
What I believe is missing is proving $A$ and $C$ are in $PQ$, I tried assuming they were not and getting to a contradiction using similar triangles but so far I have gotten nowhere. Help is appreciated.
Take a projection $\phi$ of $RF$ on $RB$ from the point $A$.
We see that $\phi(\{E\})=\{D\},$ $\phi(\{F\})=\{B\}$ and $\phi(\{R\})=\{R\}.$
Now, let $\phi(\{P\})=\{Q'\}.$
Thus, $$\frac{RP}{EP}:\frac{RF}{EF}=\frac{RQ'}{DQ'}:\frac{RB}{DB},$$ which gives $$[R,D;Q',B]=[R,D;Q,B],$$ which says $$Q'\equiv Q$$ and $A$, $P$ and $Q$ are placed on the same line.
Can you end it now?