Let $\triangle ABC$ be a triangle with incenter $I$ and orthocenter $H$. The incircle of $\triangle ABC$ touches $BC$, $CA$, $AB$ at $D, E, F$ respectively. Let $D'$ be the reflection of $D$ through $I$ and let $S$ be the midpoint of $AI$. How can we prove that $SD'$, $EF$and $HI$ are concurrent?
I'll prove the statement using complex coordinates so I'll denote all points with small letters, sorry. I won't elaborate every single detail but you should still be able to follow the proof.
Put the triangle into a complex plane and, WLOG, assume that the radius of the inscribed circle is equal to 1. Choose point $i$ as the origin, with real axis along the line $id$ (see picture).
Draw point $v$ symmetric to $d'$ with respect to $s$. Because $as=si$, $sd'=sv$ and $\angle asv=\angle isd'$ triangles $\triangle asv$ and $\triangle isd'$ are congruent (SAS). Consequentially $av=id'=1$
As a consequence $\angle sav=\angle sid'$ so lines $av$ and $id'$ must be parallel. Line $d'i$ is perpendicular to $bc$ so line $av$ must be perpendicular to $bc$ too, thus passing through orthocenter $h$.
Taking all that into account, it means that triangles $\triangle ud'i$ and $\triangle uvh$ are similar.
To finish the proof we need two lemmas and I won't prove them in detail:
Lemma 1:
$$a=\frac{2ef}{e+f}$$
You can find the proof here on page 5. Even without a proof you should be able to figure it out by yourself.
Lemma 2:
$$h=\frac{2(d^2e^2+e^2f^2+f^2d^2+def(d+e+f))}{(d+e)(e+f)(f+d)}$$
The proof can be found here, on page 4.
So we have the following coordinates:
$$i=0\tag{1}$$ $$d'=-1\tag{2}$$ $$v=a+1=\frac{2ef}{e+f}+1\tag{3}$$ $$h=\frac{2(d^2e^2+e^2f^2+f^2d^2+def(d+e+f))}{(d+e)(e+f)(f+d)}\tag{4}$$
Because triangles $\triangle ud'i$ and $\triangle uvh$ are similar:
$$\frac{v-u}{d'-u}=\frac{h-u}{i-u}\tag{5}$$
Replace (1), (2), (3), (4) into (5) and solve for $u$:
$$u=\frac{e^2 f^2+e^2 f+e^2+e f^2+e f+f^2}{e^2 f+e f^2+2 e f+e+f}\tag{6}$$
The points $u$, $e$ and $f$ are collinear if and only if:
$$\lambda=\frac{u-e}{f-u}\in R\tag{7}$$
...or, if you replace (6) into (7):
$$\lambda=\frac{(e+1) f \left(e^2-f\right)}{e (f+1) \left(e-f^2\right)}\in R\tag{8}$$
The trick is to prove that (8) is a real number for any $e,f$ from the unit circle. This is equivalent to proving:
$$\lambda - \bar\lambda=0$$
or:
$$\frac{(e+1) f \left(e^2-f\right)}{e (f+1) \left(e-f^2\right)} - \frac{(\bar e+1) \bar f \left(\bar e^2-\bar f\right)}{\bar e (\bar f+1) \left(\bar e-\bar f^2\right)}=0$$
...which leads to:
$$e^3 \bar{e}^2 f \bar{f}+e^3 \bar{e}^2 f+e^3 (-\bar{e}) f \bar{f}^3-e^3 \bar{e} f \bar{f}^2-e^2 \bar{e}^3 f \bar{f}\\ -e^2 \bar{e}^3 \bar{f}+e^2 \bar{e}^2 f-e^2 \bar{e}^2 \bar{f}-e^2 \bar{e} f \bar{f}^3+e^2 \bar{e} \bar{f}^2\\ +e^2 f \bar{f}^2+e^2 \bar{f}^2+e \bar{e}^3 f^3 \bar{f}+e \bar{e}^3 f^2 \bar{f}+e \bar{e}^2 f^3 \bar{f}-e \bar{e}^2 f^2\\ -e \bar{e} f^3 \bar{f}^2+e \bar{e} f^2 \bar{f}^3-e f^3 \bar{f}^2-e f^2 \bar{f}^2-\bar{e}^2 f^2 \bar{f}-\bar{e}^2 f^2\\ +\bar{e} f^2 \bar{f}^3+\bar{e} f^2 \bar{f}^2=0\tag{9}$$
It looks pretty hopeless, but for points $e$ and $f$ on the unit circle:
$$e\bar e = f \bar f =1\tag{10}$$
Replace (10) into (9) and you will prove that (9) is indeed zero. In other words $\lambda - \bar\lambda=0$ for any $e,f$ so $\lambda$ must be a real number. Therefore points $e,f,u$ are collinear and lines $sd'$, $hi$ amd $ef$ are congruent.