jmc

We can easily see that $△ABG\sim △ACF\sim △ADE.$

First of all, $BD=AD-AB.$ Since $AB=\frac{1}{4}AD,$ we have that $BD=\frac{3}{4}AD.$ Since $BD$ is also $DC+CB=15,$ we see that $AD=20$ and $AB=5.$ Now, we can easily find $ED=\frac{4}{5}AD=16.$

Now, we see that $CA=CB+BA=8+5=13.$ Since $\frac{FC}{CA}=\frac{ED}{DA},$ thanks to similarity, we have $FC=\frac{ED\cdot CA}{DA}=\frac{16\cdot 13}{20}=\boxed{10.4}.$