Why are these three lines concurrent?

Question

Consider a triangle $ABC$ with incentre $I$ and let $AI \cap BC=D$. Let the incentres of $\triangle ACD$ and $\triangle ABD$ be $E$ and $F$ respectively.

Prove that $AD$, $BE$ and $CF$ are concurrent.

This is part of a larger problem I am trying to solve and while working on it I noticed that this seemed true both by my diagram and on a (sort of) intuitive level.

Indeed once I placed this in Geogebra I saw this

Answer 1

Note that $E$ and $I$ both lie on the bisector of $\angle C$, so $C$, $E$ and $I$ are collinear. Similarly $B$, $F$ and $I$ are collinear.

Since $F$ lies on the angle bisector of angle $\angle BAI$, we have $\frac{|BF|}{|FI|} = \frac{|AB|}{|AI|}$ by the angle bisector theorem. In the same way, we have $\frac{|IE|}{|EC|} = \frac{|AI|}{|AC|}$. By the angle bisector theorem in $\triangle ABC$ we have $\frac{|CD|}{|DB|} = \frac{|AC|}{|AB|}$. This entails that $$ \frac{|BF|}{|FI|} \frac{|IE|}{|EC|} \frac{|CD|}{|DB|} = \frac{|AB|}{|AI|} \frac{|AI|}{|AC|} \frac{|AC|}{|AB|} = 1. $$ Since $D$, $E$ and $F$ lie on the interior of the sides of triangle $\triangle BIC$, Ceva's theorem now implies that $BE$, $CF$ and $DI$ are concurrent.