Given a tangential quadrilateral $ABCD$ with incenter $O$. Let $I, K$ be the midpoints of $AC$ and $BD$ respectively. Prove that $I, O, K$ are collinear.
This line is called the Newton line, as many people know. I found this problem difficult and have attempted to search for it online.
On this page, there is a proof of the converse of Léon Anne's theorem, which states that if given a point $O'$ lies on $IK$ (on the figure above), $S(O'AB)+S(O'CD)=S(O'BC)+S(O'DA)=\frac{1}{2}S(ABCD)$.
Area-based proving is not common in my country, as it is not easy to look out for those triangles to find correct areas. However, I did manage to prove that for this figure,$S(OAB)+S(OCD)=S(OBC)+S(ODA)=\frac{1}{2}S(ABCD)$.
All I need to finish here is to prove the normal Léon Anne's theorem, which states that:
Given a quadrilateral $ABCD$ that is not a trapezoid. Let $I, K$ be the midpoints of $AC$ and $BD$ respectively. Prove that if $O$ is a point inside the quadrilateral that satisfies $S(OAB)+S(OCD)=S(OBC)+S(ODA)=\frac{1}{2}S(ABCD)$, then $I, O, K$ are collinear.
As stated above, I don't like using area-based proving, but any helps of proving this (even if using the area-based method) will be highly appreciated.
For non-area-based proving, I think the Gauss line for complete quadrilaterals may have relation to this, as they both put midpoints on the diagonals (the Gauss line is commonly known as Newton-Gauss line).
Here's a relatively simple albeit bashy solution:
Impose the problem on the complex plane so that incircle of the quadrilateral is the unit circle, and let the tangency points of $AB,BC,CD,$ and $DA$ with the incircle be $e,f,g,$ and $h$, respectively. Then
$$A=a=\frac{2he}{h+e},\ B=b=\frac{2ef}{e+f}\ C=c=\frac{2fg}{f+g}\ D=d=\frac{2gh}{g+h}.$$
So,
$$O=0,\ I=\frac{he}{h+e}+\frac{fg}{f+g},\ K=\frac{ef}{e+f}+\frac{gh}{g+h}.$$
It suffices to prove that the value of $I/H$ is real. This value is
$$\frac{\frac{he}{h+e}+\frac{fg}{f+g}}{\frac{ef}{e+f}+\frac{gh}{g+h}}=\frac{\frac{efh+egh+efg+efh}{(h+e)(f+g)}}{\frac{efh+egh+efg+efh}{(e+f)(g+h)}} = \frac{(e+f)(g+h)}{(e+h)(f+g)}.$$
The conjugate of this value is thus
$$\frac{\left(\frac{1}{e}+\frac{1}{f}\right)\left(\frac{1}{g}+\frac{1}{h}\right)}{\left(\frac{1}{e}+\frac{1}{h}\right)\left(\frac{1}{f}+\frac{1}{g}\right)} = \frac{\frac{(e+f)(g+h)}{efgh}}{\frac{(e+h)(g+f)}{efgh}}= \frac{(e+f)(g+h)}{(e+h)(f+g)},$$
and it is thus real.