So I have rectangle $ABCD$ with isoceles triangle $EFC$ inscribed inside it.
I know the angle $\angle ECF$ and the length of $FB$ and $FC$. Is there some way to find $EA$ and $FA$? My intuition says yes, but after bashing at this problem for 30 minutes, I was still unable to find a solution. Is the problem soluble, and if so, what is the length of $EA$ and $FA$ in relation to angle $\angle ECF$, $FB$ and $FC$?
Note: I came across this problem when working on an implementation of is-point-inside-rectangle problem in CS. Being an active member of the Chem.SE, it seems that homework problems will be closed if I don't show an adequate amount of effort on the problem. If I need to edit the problem to include the work that I tried, please inform me.
With $FB$ and $FC$ you can find $\angle BCF = \arcsin\left(\frac{FB}{FC}\right)$ , which added to $\angle ECF$ gives you $\angle BCE$. Then, you can find $AB = FC \cdot\sin(\angle BCE)$ and $DE = FC \cdot\cos(\angle BCE)$.