$\text{Solution to Problem 34:}$

Let, $\bigtriangleup DEF$ be the intouch triangle. $M$ be the midpoint of $AI$ and $D_1$ be the antipode of $D$ with respect to the incircle. The Feuerbach point be $F_1$. $P,Q$ be the midpoints of the two arcs of the incircle created by the points $E,F$.

$\text{Lemma 1:}$ $F_1,D_1,M$ are collinear.

$\text{Proof:}$ From the Problem 10 of this Geometry Marathon.

$\text{Lemma 2:}$ $AI$ goes through $P,Q$

$\text{Proof:}$ Trivial.

$\text{Lemma 3:}$ $OI$ is the $\text{Euler line}$ of $\bigtriangleup DEF$.

$\text{Proof:}$ From the lemma 2 of the Solution of Problem 32 in this Geometry Marathon.

Now, we use complex numbers.

We take the incircle as the unit circle. Let the complex number of $F_1$ be $x$. Then, by using the first two lemmas and intersection formula on the two chords $F_1C$ and $PQ$, we solve for $x$=$\frac{de+ef+fe}{d+e+f}$.

Then, we are only left to show that the reflection of $F_1$ under $DE,EF,FD$ lies on the $\text{Euler line}$ from the lemma 3. Since this is symmetric, it is enough to show for just one. We get the complex number of the reflection of $F_1$ under $EF=e+f-ef (\frac{d+e+f}{de+ef+fd})=\frac{def+e^2d+f^2d}{de+ef+fd}$. So, we're left to show that $\frac{def+e^2d+f^2d}{(de+ef+fd)(d+e+f)}$ is a real number which can be shown easily. And we're done.