I'm going to write an answer using vectors.
Let $O$ be the intersection point of $AC, BD$.
Let $$\vec{OA}=\vec{a}, \vec{OB}=\vec{b}, \vec{OC}=k\vec{a}, \vec{OD}=l\vec{b}$$
where $k,l\lt 0.$
Letting $E,F,G,H$ be the midpoints of $AB, BC, CD, DA$ respectively, we have
$$\vec{OE}=\frac{1}{2}\vec a+\frac 12\vec b,\vec{OF}=\frac 12\vec b+\frac k2\vec a, \vec{OG}=\frac k2\vec a+\frac l2\vec b, \vec{OH}=\frac 12\vec a+\frac l2\vec b.$$
Letting $I$ be the intersection point of $EG, FH$, there exist $m,n$ such that
$$\vec{EI}=m\vec{EG}, \vec{FI}=n\vec{FH}.$$
The former gives us
$$\vec{OI}-\vec{OE}=m\left(\vec{OG}-\vec{OE}\right)\iff \vec{OI}=(1-m)\vec{OE}+m\vec{OG}=\frac{1-m+mk}{2}\vec a+\frac{1-m+ml}{2}\vec b.$$
The latter gives us
$$\vec{OI}-\vec{OF}=n\left(\vec{OH}-\vec{OF}\right)\iff \vec{OI}=(1-n)\vec{OF}+n\vec{OH}=\frac{k-kn+n}{2}\vec a+\frac{1-n+nl}{2}\vec b.$$
Now since $\vec a$ and $\vec b$ are linearly independent, the following has to be satisfied :
$$\frac{1-m+mk}{2}=\frac{k-kn+n}{2}\ \text{and} \frac{1-m+ml}{2}=\frac{1-n+nl}{2}.$$
These give us $m=n=1/2$ since $(k,l)\not=(-1,-1).$
Hence, we get
$$\vec{OI}=\frac{k+1}{4}\vec a+\frac{l+1}{4}\vec b.$$
On the other hand, letting $P,Q$ be the midpoints of $AC, BD$, we have
$$\vec{OP}=\frac{k+1}{2}\vec a, \vec{OQ}=\frac{l+1}{2}\vec b.$$
Finally, we obtain
$$\vec{PI}=\frac 12\vec{PQ}.$$
Since this represents that $I$ is on the line $PQ$, we now know that we get what we want. Q.E.D.
P.S. If $(k,l)=(-1,-1)$, then $ABCD$ is a parallelogram, which is an easy case.
Let the triangle be $ABC$ and external angle bisector of $\angle ABC$ cut $AC$ in $X$, of $\angle ACB$ cut $AB$ in $Y$, of $\angle BAC$ cut $BC$ in $Z$.
By angle bisector theorem,
$$\frac{AX}{XC}=-\frac{AB}{BC}... (1)$$
$$\frac{CZ}{ZB}=-\frac{CA}{AB}... (2)$$
$$\frac{BY}{YA}=-\frac{BC}{CA}... (3)$$
(1) ×(2) ×(3) gives,
$$\frac{AX.CZ.BY}{XC.ZB.YA}=-1$$
Therefore by converse of Menelaus Theorem X, Y, Z are collinear.
Best Answer
Yes, there is other method: analytic geometry.
Points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are collinear if and only if
$$\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 & 1\\x_3 & y_3 & 1\\\end{vmatrix}=0 $$
Using it we can easily solve the question:
For instance,we can put the origin of oblique cartesian axes at point B such that B, A, E lie in y-axis and B, C, F lie in x-axis. Then B has coordinates $B=(0,0)$ and without loss of generality the coordinates of A, C, D, E, F can be $A=(0,2a),C=(2c,0),D=(2d_1,2d_2),E=(0,2e),F=(2f,0)$.
As points C, D, E are collinear, we get
$$\begin{vmatrix}2c & 0 & 1\\2d_1 & 2d_2 & 1\\0 & 2e & 1\\\end{vmatrix}=0, $$ $$\begin{vmatrix}c & 0 & 1\\d_1 & d_2 & 1\\0 & e & 1\\\end{vmatrix}=0. $$
On the other hand, points A, D, F are also collinear, then
$$\begin{vmatrix}0 & 2a & 1\\2d_1 & 2d_2 & 1\\2f & 0 & 1\\\end{vmatrix}=0, $$ $$\begin{vmatrix}0 & a & 1\\d_1 & d_2 & 1\\f & 0 & 1\\\end{vmatrix}=0. $$
Summing both determinants, we get
$$c(d_2-e)+a(f-d_1)+ 1(ed_1-fd_2)=0,$$ $$\begin{vmatrix}c & a & 1\\d_1 & d_2 & 1\\f & e & 1\\\end{vmatrix}=0. $$.
Therefore the points $(c,a),(d_1,d_2), (f,e)$, aka midpoints of AC, BD, EF, are collinear.
QED.