Let $ABCD$ be a quadrilateral. $BA$$\cap$ $CD$=$K$, $BC$$\cap $$AD$=$M$. $O$ is the intersection of the diagonals. The segment parallel to $MK$ and goes through $O$ meet at sides $BC$ and $AD$ at $E$ and $F$, respectively. Then prove that $EO$=$OF$.
My try: Let $BF$$\cap $$MK$=$P$ and $BD$$\cap$ $MK$=$Q$. From here, if we can prove $P, D, E$ are collinear, we can say $MQ=QP$ by the Ceva theorem. Moreover it will be $EO$=$OF$. However, I can't prove it at all. Can you guys give me some hints?
I don't know any complex, vector and projective geometry, so please consider giving me an Euclidean one.
Instead of $P=BF\cap MK$, let $P=BF\cap ED$. Then use Desargues' Theorem with triangles $AFB$ and $CED$ to show $P$ is on line $MK$.
Complete the proof as you suggest, with Ceva's theorem.
Desargues' Theorem is traditionally a theorem of projective geometry, but it can be easily proved using Menalaus' Theorem, as shown on page 2 of the reference above.