Let $ABC$ be a triangle. Let $I$ be its inscribed circles' center. Let $D$ be the intersection point of lines $AI$ and $BC$. The perpendicular bisector of $AD$ intersects with $BI$ in $E$.
Thesis: $$ ED\perp CI $$
How to prove that using synthetic euclidean geometry?
The key is to show that $A,B,D,E$ are cyclic. To show this let the angle bisector of $\angle B$ cut the circumcircle of $\triangle ABD$ at $E^{'}$. Then, since $\angle ABE^{'}=\angle DBE^{'}$, we have $AE^{'}=DE^{'}$. Therefore the perpendicular bisector of $AD$ passes through $E^{'}$. Hence $E^{'}$ is the same as $E$, that is $A,B,D,E$ are cyclic as we desired. The remaining part is trivial.
Let $P$ be the intersection of $DE$ and $CI$ and $F$ the midpoint of $AD$. Using the cyclicity we've just proven, $$\angle AED=\pi-\angle ABD=2(\angle DAC+\angle ICA)$$ Because$ \triangle AED$ is isosceles, it's obvious that $\square PEFI$ is cyclic and the conclusion follows.
