Belarusian Mathematical Olympiad

Answer: $2:5$. Let the segment $MD$ intersect the side $AC$ at $N$ and the diagonal $AC$ at $P$. From $BC\parallel AD$ we get $\angle CBK=\angle BKA$. Since $BK$ is the bisector, $\angle ABK=\angle CBK$ hence $\angle ABK=\angle BKA$ and the triangle $ABK$ is isosceles with $AB=AK$. From $2AB=AD$ it follows that $K$ is the midpoint of $AD$, similarly $N$ is the midpoint of $BC$. It is clear that $BNDK$ is a parallelogram, therefore $ND=BK$ and $NP$, $LK$ are the middle lines of the triangles $BCL$ and $APD$ respectively. Thus $2NP=BL$ and $2LK=PD$ whence $NP+PD=ND=BK=BL+LK$ and therefore $2NP=2LK=PD$. Since $N$ is the midpoint of $BC$, $BN=0.5BC=0.5AD$. And since $BC\parallel AD$, $BN$ is the middle line of the triangle $AMD$, so $N$ is the midpoint of $MD$, i.e.



$$MN=ND=NP+PD=\frac{3}{2}PD⟹MP=MN+NP=\frac{3}{2}PD+\frac{1}{2}PD=2PD.\text{\hspace{1em}(1)}$$ Let $Q$ be the intersection point of the side $AD$ and the line passing through $P$ parallel to $MF$. By Thales' theorem $FQ:QD=MP:PD\stackrel{(1)}{=}2:1$. Since $LK$ is the middle line of the triangle $APD$, $AK=KD$. Therefore $$AF:AD=AF:(AF+FQ+QD)=FQ:(2FQ+QD)=2QD:5QD=2:5.$$