Let $M$ be the projection of $E$ on $CB$ then the triangles $EMC, EFC$ are congruent $\rightarrow CF=CM=\frac{CB} {2}$.

Let S be a midpoint of BC. Then <DCF = <DBC = <BCA, CE - bisector of angle ACD => <SCE = <FCE and <SEC = 90 - <SCE = <FEC => triangles SCE and FCE are congruent => SC = CF = BC/2.

If $M$ is the midpoint of $BC$ and $\angle ECF = \alpha$, we know $EM$ is perpendicular to $BC$ and $\angle EBC = \alpha \Rightarrow \angle MEC = \angle BEM = 90-\alpha = \angle CEF$. Thus by $HA$ congruency, $\triangle MEC \cong \triangle FEC \Rightarrow CF=MC \Rightarrow BC=2CF$.