i) Any such quadrilateral $ABCD$ can be obtained from some triangle $AFE$ by drawing a cevian $DB$ ($D$ on $AF$ and $B$ on $AE$) parallel to $FE$ and joining $B$ to $F$ and $D$ to $E$, letting the intersection of these two segments be $C$. So this part of the problem reduces to showing that for any such cevian $DB$ in any triangle $AFE$, the midpoint $G$ of $FE$ lies on $AC$ extended. ii) By above, this is equivalent to showing that in any triangle $AFE$ and any point $C$ on $AG$, where $G$ is the midpoint of $FE$, then if the intersection of $FC$ past $C$ with $AE$ is $B$ and the intersection of $EC$ past $C$ with $AF$ is $D$, then $DB$ is parallel to $FE$.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(13); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -24.498946978935827, xmax = 9.737361310785264, ymin = -4.690889826670544, ymax = 15.654457312711958; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); draw((-8.3,3.91)--(-1.28,5.13)--(2.18,-3.25)--(-9.871990625049701,-1.0196999847606336)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-8.3,3.91)--(-1.28,5.13), linewidth(2) + rvwvcq); draw((-1.28,5.13)--(2.18,-3.25), linewidth(2) + rvwvcq); draw((2.18,-3.25)--(-9.871990625049701,-1.0196999847606336), linewidth(2) + rvwvcq); draw((-9.871990625049701,-1.0196999847606336)--(-8.3,3.91), linewidth(2) + rvwvcq); draw((-22.848332541718666,1.3816573075645637)--(-5.021403232613448,14.191548869740084), linewidth(1.2)); draw((-22.848332541718666,1.3816573075645637)--(-9.871990625049701,-1.0196999847606336), linewidth(1.2)); draw((-22.848332541718666,1.3816573075645637)--(-8.3,3.91), linewidth(1.2)); draw((-1.28,5.13)--(-5.021403232613448,14.191548869740084), linewidth(1.2)); draw((-5.021403232613448,14.191548869740084)--(-8.3,3.91), linewidth(1.2)); draw((-9.871990625049701,-1.0196999847606336)--(-1.28,5.13), linewidth(1.2)); draw((-22.848332541718666,1.3816573075645637)--(-0.09700775518795428,2.2648338116979936), linewidth(0.8) + linetype("4 4")); draw((2.18,-3.25)--(-13.954005709535478,7.77285122903378), linewidth(1.2)); /* dots and labels */ dot((-8.3,3.91),dotstyle); label("$A$", (-8.699732579662898,4.4855632968854255), NE * labelscalefactor); dot((-1.28,5.13),dotstyle); label("$B$", (-0.8141566539156627,5.187126991346892), NE * labelscalefactor); dot((2.18,-3.25),dotstyle); label("$C$", (2.4130363406072273,-3.905138488873703), NE * labelscalefactor); dot((-9.871990625049701,-1.0196999847606336),dotstyle); label("$D$", (-10.439610541927411,-1.7723848577108472), NE * labelscalefactor); dot((-22.848332541718666,1.3816573075645637),linewidth(4pt) + dotstyle); label("$E$", (-23.57288290224665,0.9496822767996393), NE * labelscalefactor); dot((-5.021403232613448,14.191548869740084),linewidth(4pt) + dotstyle); label("$F$", (-4.91128862957081,14.58808049713053), NE * labelscalefactor); dot((-13.954005709535478,7.77285122903378),linewidth(4pt) + dotstyle); label("$G$", (-14.3683672309118,8.302069794755798), NE * labelscalefactor); dot((-5.580431945061905,2.0519744968171025),linewidth(4pt) + dotstyle); label("$M$", (-5.865415254038447,2.689560239064074), NE * labelscalefactor); dot((-0.09700775518795428,2.2648338116979936),linewidth(4pt) + dotstyle); label("$X$", (0.47672054389349316,2.2686220223871945), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

By Ceva's theorem we have: $\frac{EM}{MF} \cdot \frac{FD}{FA} \cdot \frac{AB}{BE} =1$ But $BD$ || $EF \implies \frac{AD}{DF} = \frac{AB}{BE} $ So $\frac{EM}{MF} =1 \implies EM=MF$ $\boxed{Q.E.D}$