In $\triangle ABC$, $AB > BC > CA, AB = 6, \angle C - \angle B = 90^{\circ}$, circle $O$ is the incircle, $E$ is the tangency point on side $BC$. $EF$ is a diameter of circle $O$. Ray $AF$ intersects side $BC$ at point $D$. If $DE$ equals the circumradius of $\triangle ABC$, find the lengths of sides $BC$ and $AC$.

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Draw the tangent line of $\odot O$ through $F$, intersecting $A B$ and $A C$ at $M$ and $N$, respectively. Clearly, $M N \parallel B C$; Draw $C K \perp B C$, with $K$ on $A B$.

It is easy to know that $\triangle A M F \sim \triangle A B D$ and $\triangle A M N \sim \triangle A B C$.

By the tangent length theorem, we have $A M + M F = A N + N F = \frac{1}{2}$ the perimeter of $\triangle A M N$.
Since $\frac{A M}{A B} = \frac{M F}{B D} = \frac{A M + M F}{A B + B D} = \frac{\frac{1}{2} \text{ the perimeter of } \triangle A M N}{A B + B D}$,

and
$$
\frac{A M}{A B} = \frac{\text{ the perimeter of } \triangle A M N}{\text{ the perimeter of } \triangle A B C}
$$

Therefore,
$$
A B + B D = \frac{1}{2} \text{ the perimeter of } \triangle A B C.
$$

According to the tangent length theorem, it is easy to prove that
$$
A B + E C = \frac{1}{2} \text{ the perimeter of } \triangle A B C,
$$

Thus,
$$
B D = E C.
$$

Let $B C = a$, $C A = b$, $A B = c$, and $p = \frac{1}{2}(a + b + c)$. 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 $\triangle A C K \sim \triangle A B C$, thus $A K = \frac{b^2}{c}$, $C K = \frac{a b}{c}$, $B K = \frac{c^2 - b^2}{c}$.
Since $2 R = \frac{b}{\sin B} = \frac{b}{\frac{C K}{B K}} = \frac{b \cdot B K}{C K}$
$$
= \frac{b \cdot \frac{c^2 - b^2}{c}}{\frac{a b}{c}} = \frac{c - b}{a}(c + b),
$$

and
$$
2 R = 2(c - b),
$$

Therefore,
$$
b + c = 2 a.
$$

From the right triangle $\triangle B C K$, we get $B K^2 = B C^2 + C K^2$, i.e., $\left(\frac{c^2 - b^2}{c}\right)^2 = a^2 + \left(\frac{a b}{c}\right)^2$ or $\frac{(c - b)^2(2 a)^2}{c^2} = a^2 + \frac{a^2 b^2}{c^2}$, simplifying to
$$
b^2 + c^2 = 4(c - b)^2,
$$

rearranging and transforming to $\left(\frac{b}{c}\right)^2 - \frac{8}{3}\left(\frac{b}{c}\right) + 1 = 0$,

solving for
$$
\frac{b}{c} = \frac{8 \pm 2 \sqrt{7}}{6}.
$$

Note that $b < c$, so $\frac{b}{c} = \frac{8 - 2 \sqrt{7}}{6}$.
Since $c = 6$, then $b = 8 - 2 \sqrt{7}$, $a = \frac{1}{2}(b + c) = 7 - \sqrt{7}$.