I have attached a .png image with a figure I have drawn in GeoGebra, it will make it easier to view the proof. First of all, there are 3 total possibilities for the point $D$ to lay on side $BC$: Either between $B$ and $C$(in my figure $F$), to the left of $B$(in my figure $D$) or to the right of $C$(in my figure $G$). Now let's show that $D$ is always on the left side of $B$. In the case of $F$, by applying the bisector theorem we get $\frac{AC}{BC} = \frac{AF}{BF}$, which follows $\frac{AF}{AC} = \frac{BF}{BC}$ , however if you consider $F' \in [AB]$ the intersection of the parallel through $F$ to $AC$ you would quickly get $\frac{BF}{BC} = \frac{FF'}{AC} = \frac{AF}{AC} \Rightarrow AF = FF'$, which is obviously false. With the same logic you can easily prove that $D$ cannot be on the left side of $BC$. So $D$ is on the left side of $B$. Since $DE$ and $CE$ are bisectors it follows that $E$ is the incircle of $\bigtriangleup{ADC}$, So $AB$ is the bisector of $\angle{DAC}$, which means that $\angle{DAB} = \angle{BAC}$.From there we can easily conclude that the symmetric of $D$ with respect to $AB$ is on the line $AC \Rightarrow D' \in AC$

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