In scalene triangle $ABC$, $AD$ and $AD'$ are internal and external bisectors of $\angle A$, respectively, meeting $BC$ at $D$ and $D'$, respectively. If $A'$ is the midpoint of $DD'$, and similarly $B'$ and $C'$ are the corresponding on $CA$ and $AB$, prove that $A'$, $B'$, and $C'$ are collinear.
I used vectors to solve this question:
Let $A\equiv \vec a$, $B\equiv\vec b$, $C\equiv\vec c$. This implies, $D\equiv\dfrac{b\vec b+c\vec c}{b+c}$ and $D'\equiv\dfrac{b\vec b-c\vec c}{b-c}$. Further, $$A' \equiv \frac {b^2\vec b-c^2\vec c}{b^2-c^2} \qquad B' \equiv \frac {c^2\vec c-a^2\vec a}{c^2-a^2} \qquad C' \equiv \frac {a^2\vec a-b^2\vec b}{a^2-b^2}$$
Therefore, $$(b^2-c^2)A'+(c^2-a^2)B'+(a^2-b^2)C'=0$$
This implies $A'$, $B'$, and $C'$ are collinear, since coefficients of $A'$, $B'$, and $C'$ add to $0$.
But is there any way to prove this by only using euclidean geometry? I can't even think of an approach to start with.
