6. Given that $C D$ is the altitude on the hypotenuse $A B$ of the right triangle $\triangle A B C$, and a perpendicular line $D E$ is drawn from point $D$ to the leg $B C$. If the segments $A C, C D, D E$ can form a triangle (i.e., they are the three sides of a triangle), then the range of $\frac{B C}{A B}$ is ( ).
(A) $\left(\frac{\sqrt{5}-1}{2},+\infty\right)$
(B) $(0,1)$
(C) $\left(\frac{\sqrt{5}-1}{2}, 1\right)$
(D) $\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$
6.C.

In Rt $\triangle A B C$,
$$
\begin{array}{l}
0A C \\
\Rightarrow A C \sin A + A C \sin ^{2} A > A C \\
\Rightarrow \sin ^{2} A + \sin A - 1 > 0 \\
\Rightarrow \left(\sin A + \frac{\sqrt{5} + 1}{2}\right)\left(\sin A - \frac{\sqrt{5} - 1}{2}\right) > 0 .
\end{array}
$$

Since $\sin A + \frac{\sqrt{5} + 1}{2} > 0$, we have,
$$
\sin A - \frac{\sqrt{5} - 1}{2} > 0 \text {, }
$$

which means $\sin A > \frac{\sqrt{5} - 1}{2}$.
From equations (1) and (2), we know $\frac{\sqrt{5} - 1}{2} < \sin A < 1$