Let $M$ be the midpoint of segment $[AB]$ in triangle $\triangle ABC$. Let $X$ and $Y$ be points such that $\angle{BAX}=\angle{ACM}$ and $\angle{BYA}=\angle{MCB}$. Both points, $X$ and $Y$, are on the same side as $C$ with respect to line $AB$.
Show that the rays $[AX$ and $[BY$ intersect on line $CM$.
Consider $BY$ intersect $CM$ at $N$ so $\triangle BMN$ and $\triangle CMB$ are similar
=> $BM^2=CM.MN$
with the same way if $AX$ intersect $CM$ at $K$
$AM^2=CM.MK$
so $MN=MK$ which gives us our result
Niosha wrote:
Consider $BY$ intersect $CM$ at $N$ so $\triangle BMN$ and $\triangle CMB$ are similar
=> $BM^2=CM.MN$
with the same way if $AX$ intersect $CM$ at $K$
$AM^2=CM.MK$
so $MN=MK$ which gives us our result
Almost correct. You still have to consider why those rays actually intersect $CM$.
I thought it's obvious. if those rays don't intersect $CM$ so for example $AX$ and $MC$ are parallel
=> $\angle{MCA}=\angle{CAX}=\angle{BAX}$
$\angle{CAX}<\angle{BAX}$ which is a contradiction