An interesting ratio

In triangle $ABC$ as shown above, the centroid $G$ is determined by the intersections of $AD$ , $BE$ and $CF$ , where $D$ , $E$ and $F$ are the midpoints of $BC$ , $AC$ and $AB$ respectively.

Now suppose $AG$ is bisected with its midpoint $M$ and $BM$ extended meets $AC$ at $K$ , while $CM$ extended meets $AB$ at $L$ . If $\dfrac{AL}{AB}=\dfrac{AK}{AC}=\frac{1}{X}$ , find $X$ .