Concurrent lines in circumscribed pentagon 
Let $A'B'C'D'E'$ be a pentagon with an incircle $\odot (I)$ touching $C'D',D'E',E'A',A'B',B'C'$ at $A,B,C,D,E$ respectively. Suppose that $AA',BB',EE'$ concur at $I$. Prove that $AA',CC',DD'$ concur too.

This is easy with complex numbers. Denote the complex number of a point by the corresponding lower case letter. Since $C,D$ are symmetric in the line $AI$, we get $a^2=cd$. Analogously, we have $b^2=de,e^2=bc$ so $be=cd=a^2$: i.e. $B,E$ are symmetric in $AA'$. Thus $CC',DD'$ are symmetric in $AA'$, so we're done. 
However, I think that I'm missing a much simpler argument relying on symmetry alone. To start with, $C,D$ are symmetric in $AA'$, $B,C$ are symmetric in $EE'$, and $D,E$ are symmetric in $BB'$. For some reason though, I cannot seem to put together a proof using symmetry only. 

Notice first of all that $\angle CIA'=\angle DIA'$, $\angle DIB'=\angle EIB'$ and $\angle CIE'=\angle BIE'$. But $\angle EIB'=\angle BIE'$ (vertical angles): it follows that $\angle EIA'=\angle BIA'$ and thus $\angle D'IA=\angle C'IA$. That means that $D'$ is the reflection of $C'$ about $AA'$.
