Prove that in a quadrilateral, the lines joining the midpoints of the opposite sides and the midpoints of the diagonals are concurrent.
We construct an arbitrary quadrilateral $ABCD$ with $E, F, G$ as the midpoints of $AB, BC, CD$. Let $H, I$ be the midpoints of $AC, BD$. Let $EG, HI$ intersect at $J$. Let the line joining $F, J$ meet $AD$ at $K$. We will prove that $K$ is the midpoint of $DA$.
Joining $KG, GF, FE, EK$, it quickly becomes clear that the above is only true if $KGFE$ is a parallelogram, which in turn, is only true if $EJ = JG, KJ = JF$. Proving the first equality is easy.
In $\Delta ABC, EH || BC, 2\cdot EH = BC$. Likewise, in $\Delta DBC, IG || BC, 2\cdot IG = BC$. Therefore, $IG||EH, IG=EH$. Therefore, $EHGI$ is a parallelogram and $EJ = JG$. Even after numerous efforts I wasn't able to prove the second equality.
I noticed that this was because I was not utilizing the fact that $F$ is the midpoint of $BC$ and that $FK$ is the straight line.
So, to utilize those facts, I considered $\Delta HKJ, \Delta JFI$. Proving these are congruent will prove our conjecture. Now, we can use the fact that $FK$ is a straight line by saying, that $\angle HJK = \angle IJF$. Also, since $EHGI$ is a parallelogram, $HJ = JI$. Now we need only one more equivalence to prove congruency. I wasn't able to find this.
A way to utilize the fact that $F$ is the midpoint of $BC$ is by noticing that $EHCF, IGFC, AHFE, FIDG$ are all parallelograms. I have, however, no idea how to use these in the proof.
I think I'm forgetting something. Because in each approach I take, there is always a single piece that is missing. If anybody could point out what this 'piece' is, I would be grateful. I would appreciate solutions that are related to the approaches described above.