In triangle $ABC$, $AB > BC > CA$, $AB=6$, $\angle{B}-\angle{C}=90^o$. The incircle touches $BC$ at $E$ and $EF$ is a diameter of the incircle. Radical $AF$ intersect $BC$ at $D$. $DE$ equals to the circumradius of $\triangle{ABC}$. Find $BC$ and $AC$.