Question 133
Let $\triangle ABC$ be a triangle in which $\angle ABC = 75^\circ$ and $\angle BAC = 45^\circ$. One considers the points $F$, $X$ and $Y$ such that $F$ is the projection of $B$ on $AC$, $\overline{CX} = \frac{1}{2}\overline{BC}$ and $\overline{BY} = \overline{FX} + \overline{FA}$. Prove that the centroid of triangle $ABX$ lies on the line segment $YF$.