We are given that in the regular hexagon with side length 1:

\vec{AB}=\mathbf{u} \quad \text{and} \quad \vec{BC}=\mathbf{v}.

First note that (see diagram above)

\vec{AE}\,=\vec{AD}+\vec{DE}=2\mathbf{v}-\mathbf{u}\quad

and

\vec{AX}=\vec{AF}+\vec{FX}

Now we know that X lies along CF such that \left|\vec{XC}\right|=3\left|\vec{FX}\right|, and \left|\vec{FC}\right|=\left|\vec{FX}\right|+\left|\vec{XC}\right|. Therefore:

\left|\vec{FC}\right|=4\left|\vec{FX}\right|

And hence \vec{FX}=\frac14\vec{FC}=\frac12\mathbf{u} . Therefore:

\vec{AX}=\vec{AF}+\vec{FX}=\left(\mathbf{v}-\mathbf{u}\right)+\frac12\bm{u=v}-\frac12\mathbf{u}

Since we want to show that the points \mathrm{A}, \mathrm{X}, and \mathrm{E} are colinear, it suffices to show that \vec{AX} is parallel to \vec{AE} (or to \vec{XE}). Notice that:

\vec{AX}=\mathbf{v}-\frac{1}{2}\mathbf{u}=\frac{1}{2}(2\mathbf{v}-\mathbf{u})=\frac{1}{2}\vec{AE}.

Since multiplying \vec{AE} by the scalar \frac{1}{2} simply scales the vector, we conclude that \vec{AX} and \vec{AE} have the same direction. Therefore, the points \mathrm{A}, \mathrm{X}, and \mathrm{E} are colinear.

Vectors