One, (50 points) As shown in the figure, in $\triangle A B C$, $A B>B C>C A, A B=6, \angle C-\angle B=90^{\circ}$, circle $O$ is the incircle of $\triangle A B C$, $E$ is the tangency point on side $B C$, and $E F$ is a diameter of circle $O$. Ray $A F$ intersects side $B C$ at point $D$. If $D E$ equals the radius of the circumcircle of $\triangle A B C$, find the lengths of sides $B C$ and $A C$.
One, Solution: Draw a tangent line through $F$ to circle $O$, intersecting $A B, A C$ at $M, N$. Clearly, $M N \parallel B C$. Draw $C K \perp B C$, with $K$ on $A B$. It is easy to see that $\triangle A M F \backsim \triangle A B D, \triangle A M N \backsim \triangle A B C$. By the tangent length theorem, we have $A M + M F = A N + N F = \frac{1}{2} \text{ perimeter of } \triangle A M N$. Since $\frac{A M}{A B} = \frac{M F}{B D} = \frac{A M + M F}{A B + B D} = \frac{\frac{1}{2} C_{\triangle A M N}}{A B + B D} \left(C_{\triangle M M N} \text{ refers to the perimeter of } \triangle A M N, \text{ same below}\right)$. (1);
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\text{Also, } \frac{A M}{A B} = \frac{C_{\triangle M M N}}{C_{\triangle A B C}}
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From (1) and (2), we get $A B + B D = \frac{1}{2} \text{ perimeter of } \triangle A B C$.
By the tangent length theorem, it is easy to prove that $A B + E C = \frac{1}{2} \text{ perimeter of } \triangle A B C$, so $B D = E C$.
Let $B C = a, C A = b, A B = c, p = \frac{1}{2}(a + b + c)$. It is easy to know that $B D = E C = p - c$, at this time the circumradius $R$ of $\triangle A B C$ is $D E = B C - (B D + E C) = a - 2(p - c) = c - b$.
It is easy to prove that $\triangle A C K \backsim \triangle A B C$, so $\frac{A C}{A B} = \frac{A K}{A C} = \frac{C K}{B C}$, thus $A K = \frac{b^2}{c}, C K = \frac{a b}{c}, B K = \frac{c^2 - b^2}{c}$.
Since $2 R = \frac{b}{\sin B} = \frac{b}{\frac{C K}{B K}} = \frac{b \cdot B K}{C K} = \frac{b \cdot \frac{c^2 - b^2}{c}}{\frac{a b}{c}} = \frac{c^2 - b^2}{a}$.
And $2 R = 2(c - b)$, so $b + c = 2a$. From the right triangle $\triangle B C K$, we get $B K^2 = B C^2 + C K^2$,
i.e., $\left(\frac{c^2 - b^2}{c}\right)^2 = a^2 + \left(\frac{a b}{c}\right)^2$, combining $b + c = 2a$ and simplifying, we get $b^2 + c^2 = 4(c - b)^2$.
$\left(\frac{b}{c}\right)^2 - \frac{8}{3}\left(\frac{b}{c}\right) + 1 = 0$, solving gives $\frac{b}{c} = \frac{8 \pm 2 \sqrt{7}}{6}$, since $b < c$, so $\frac{b}{c} = \frac{8 - 3 \sqrt{7}}{6}$.
Given $c = 6$, thus $b = 8 - 3 \sqrt{7}, a = \frac{b + c}{2} = 7 - \sqrt{7}$.