In quadrilateral $ABCD$, let $AB$ and $CD$ meet at $E$ and $AD$ and $BC$ meet at $F$. Then prove that the midpoints of $AC,BD$ and $EF$ are collinear.
What I Tried:- Here is a Picture :-
First extend $GH$ to meet $CE$ at $J$. Now I applied Menelaus Theorem on whatever triangles I found suitable :-
$$\rightarrow \frac{BI}{ID} * \frac{DJ}{CJ} * \frac{CL}{BL} = 1.$$ $$\rightarrow \frac{AK}{KD} * \frac{DJ}{CJ} * \frac{CH}{HA} = 1.$$ $$\rightarrow \frac{FG}{GE} * \frac{EJ}{CJ} * \frac{CL}{LF} = 1.$$ $$\rightarrow \frac{FG}{GE} * \frac{EJ}{DJ} * \frac{DK}{KF} = 1.$$
From here I got some results, like $\frac{AK}{KD} = \frac{CL}{BL}$ , but I do not know how to proceed further.
Can someone help me? Thank You.
