Let $AD$ intersect $CE$ at $X$ then $\frac{CX}{XE}=\frac{[ACD]}{[AED]}=\frac{AC\cdot CD}{AE\cdot ED}$. Thus $AD,BE,CF$ are concurrent iff $\frac{AC\cdot CD}{AE\cdot ED}\cdot \frac{CE\cdot EF}{CA\cdot AF}\cdot\frac{EA\cdot AB}{EC\cdot CB}=\frac{AB}{BC}\cdot \frac{CD}{DE}\cdot\frac{EF}{FA}=1$.

For this problem use Ceva theorem for cyclic convex hexagon

notice that :$AB=2Rsin( ADB)$ and similary for the other lengths ,where $R$ is the circumraduis.Thus $$\frac {AB}{BC}\cdot\frac {CD}{DE}\cdot\frac {EF}{FA}= 1$$if and only if $$\frac{sinFDA}{sinADB}. \frac{sinDBE}{sinEBF}.\frac{sinBFC}{sinCFD}=1$$which is ceva trigonometrique form ie the lines $AD$,$BE$ and $CF$ are concurrent.

Can you use similar triangles to solve this?