Let $ABC$ be a triangle where $AC \geq AB$. Let $C_1$ be the midpoint of $AB$ and $B_1$ be the midpoint of $AC$. Let $D$ be the intersection of:
- The line perpendicular to the internal bisector of $C$ that passes through $C_1$.
- the line perpendicular to the internal bisector of $B$ passing through $B_1$.
Let the line passing through $D$ that is parallel to the internal bisector of $A$ intersect side $AC$ at $E$. Lastly, let $F$ be the intersection of the perpendicular bisector of $BC$ with the arc of the circumcircle of $ABC$ containing $A$. Show that $EF \perp AC$.
I tried using Archimedes' theorem (regarding cleavers), but that does not give much. I also tried to consider the medial triangle of $ABC$ and its circumcircle. With it, Archimedes' theorem, we know that $DE$ is the angle bisector of one of the angle of the medial triangle of $ABC$.
I was able to reduce this to showing that DE passes through the midpoint of BC.