Prove that the $AX$ bisects $BC$ in $\triangle ABC$ where $X$ is the intersection of a side $EF$ of the contact triangle $DEF$ and $ID$

Question

In Evan Chen's Euclidean Geometry in Mathematical Olympiads, the title of the question was reduced to proving $AX$ bisecting (using homothety) in Problem 4.16 ( Page $63$):

Prove that $XB'=XC'$

Where $DEF$ is the contact triangle of $ABC$, ray $ID$ is drawn perpendicular to $BC$ and intersects $FE$ at $X$ ,and $B'C'$ is a line segment drawn parallel to $BC$ through $X$

However, I am unable to tackle this (Even with the hint: Note that AI bisects $∠B'AC'$)

Any extra hint or solution would be appreciated, thanks a lot.

Answer 1

Note that $IXB'F$ and $IC'EX$ are cyclic and that triangle $EFI$ is isosceles ($FI = EI$). So we have $$\angle XB'I = \angle XFI = \angle XEI = \angle XC'I$$ so triangle $B'C'I$ is also isosceles and thus conclusion.