It is given to us that $ABCD$ is a parallelogram. $E$ and $F$ are mid- points of $CD$ and $BC$ respectively. $AC$ and $BD$ are it's diagonals. We have to prove that $AC$ bisects $GH$.
One may start by observing that $AC$ and $GH$ are diagonals of the quadrilateral $AHCG$. So proving it a parallelogram solves the problem. However, when I tried to prove it by using congruence, I ended up with one equality less. Similarity didn't work either. Can I get some hints for solving this?
I tried to prove that $AHCG$ is a parallelogram by proving $\Delta IGC \cong \Delta IHA$. The relations are $$IA = IC$$ $$\angle AIH = \angle CIG$$
However, there is no third relation. If I somehow prove that $GC = HA$ i can prove that $AHCG$ is a parallelogram. But I can't seem to get any new relation.
P.S. - I don't have a ton of knowledge in mathematics. An answer using elementary math would be appreciated.