Let $ABCD$ be a quadrilateral with the midpoints of every side $E,F,G,H.$ $AG,DE$ meet at $P$ and $CE,BG$ meet at $Q$. Show that $PQ|| HF$.
Indeed, we can assert that $R$, the intersection point of $AC,BD$, lies on $PQ$, by Pappus theorem. This will help?
Please give a pure geometric proof rather than algebraic one.
Now, I come up with a proof based on projective geometry as follows. Please correct me if I'm wrong.
First, let's introduce a theorem, which belongs to Issac Newton.
Theorem 1 The midpoints of three diagonals of a complete quadrilateral are collinear.
To be specific, as Figure 1 shows, $G,H,I$ are the diagonals $AD,BF,CE$ respectively of a complete quardrilateral $ABCDEF$, then $G,H,I$ are collinear. The line is called Newton line.
By theorem 1, we can further obtain a conclusion as follows:
Corollary 1 Based on Figure 1, as Figure 2 shows, construct $CJ \parallel AE, EJ \parallel AC$, then $DJ \parallel HI.$
Indeed, we have $A,I,J$ are collinear, hence $GI$ is a median line of $\triangle AJD$. Therefore, $DJ \parallel GI$. But by theorem 1, $H$ is also on $GI$. It follows that $DJ \parallel HI$.
Now, let's derive a more profound theorem, which is
Theorem 2 As Figure 3 shows, take $K,L$ on $BC,FE$ respectively such that $\dfrac{BK}{KC}=\dfrac{FL}{LE}=k$. $BL$ meets $ FK$ at $M$, and $CL$ meets $EK$ at $N$. Then $M,N,J$ are collinear.
We intend to prove at the standpoint of projective geometry. Notice that the two crossratioes are equal, i.e. $(B,K,C,\infty)=(F,L,E,\infty)$, where $\infty$ denotes the point at infinity. Thus, the $EB,EK,EC,EJ$ and $CF,CL,CE,CJ$ are projectively corresponding. Specially, $EC$ corresponds to itself. Therefore, this is also a perspectivity, which implies the intersection points $D,N,J$ of the corresponding lines are collinear. But by Pappus theorem, $M,D,N$ are collinear. Then $M,D,N,J$ are collinear. By Corollary 1, $ND \parallel IH$.
Finally, let's go back to the preasent problem. It's obvious that, it is just the special case when $k=1$. We are done.
Given any unequal segments of two non-parallel lines, $BK$ and $CJ$ divided into any equal number of parts, join corresponding points $CB$, $GE$, $DA$, $LN$, $JK$. I assume that if $CB$ and $JK$ are not parallel, then none of the joined lines are parallel.
I assume also, that if $F$ and $M$ bisect $CB$ and $JK$, then $FM$ bisects $GE$, $DA$, $LN$ at $O$, $H$, $T$.
Draw the diagonals of each of these non-parallel-sided quadrilaterals to get points $Q$, $P$, $R$, $S$, and join each pair of successive points with $QP$, $PR$, $RS$.
$ABCD$ is thus the quadrilateral given in the question, rotated $90^o$ clockwise for convenience, and with $GE$ joined. It seems clear that, if lines $CB$, $GE$, $DA$ are pairwise convergent somewhere above $BK$, as in the figure, then points $Q$ and $P$, where the pairs of diagonals intersect, and hence the segment $QP$, lie above the bi-medial line $FH$.
Similarly, in quadrilateral $NEGL$, $P$ and $R$, where the pairs of diagonals intersect, and hence segment $PR$, lie above bi-medial $GT$; and in quadrilateral $KADJ$, $RS$ lies above bi-medial $HM$.
It may not be safe to assume, however, that $QP$, $PR$, $RS$...are collinear. But if, as OP suggests, we know, from a theorem of Pappus or otherwise, that the intersection $I$ of diagonals $AC$, $BD$ lies on $QP$, and likewise intersection $U$ of diagonals $NG$, $EL$ lies on $PR$ and further conversely, that $P$ lies on $IU$ and $R$ on $UV$, then the overlapping segments $QP$, $IU$, $PR$, $UV$, $RS$ are indeed collinear. And a line which lies on the same side of another line however far they are extended, is parallel to that line.
Therefore$$QP\parallel FH$$
Note: Things get a little chaotic where $BK$ and $CJ$ intersect. When the upper base of the quadrilateral lies partly above and partly below the lower, or vice-versa, the quadrilateral is concave, but the result still follows; and after that, the sides of the convex quadrilaterals again converge in the same direction, with the intersection points of the pairs of diagonals lying beyond the bi-medial line in the direction of convergence, as in the figure below.
$NADL$ and $KNLJ$ are concave quadrilaterals, with diagonals $LA$, $DN$ and $JN$, $LK$ meeting externally at $R$ and $S$, respectively.
Then we are back to normal with quadrilaterals $IKJV$, $ZIVW$, etc.
