Consider a 2-simplex, represented as the equilateral triangle abc below. Since $ABC$ is a simplex, we also know that the distance from each vertex to the opposite side is 1.
The line $\overline{DG}$ intersects $\overline{AB}$ perpendicularly (similarly for $\overline{GE}$ and $\overline{BC}$). I know the length of $\overline{DG}$, call it $x$, and the length of $\overline{GE}$, call it $y$. Can the lengths of $\overline{AF}$ and $\overline{FC}$ be stated in terms of $x$ and $y$?
By Law of sines: $$\begin {cases} \frac {\sin BCF}{BF}=\frac {\sin FBC}{FC}\\ \frac {\sin BAF}{BF}=\frac {\sin ABF}{AF}\\ \angle BAF=\angle BCF \end {cases}\Rightarrow\frac {AF}{FC}=\frac {\sin ABF}{\sin FBC}=\frac {x}{y}. $$ The same result could be obtained by drawing from the point $F$ lines parallel to $GD$ and $GE$, respectively, and considering resulting similar triangles.
Can you proceed further?