6. Given that $A D$ is the altitude of $\triangle A B C$, $B C+A D-A B$ $-A C=0$. Find the range of $x=\angle B A C$.

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Solution: Clearly, $A D \leqslant A B, A D \leqslant A C$. Therefore, from $B C + A D - A B - A C = 0$, we get $B C \geqslant A C, B C \geqslant A B$. Thus, $\angle B, \angle C$ are both acute angles.

If $\angle A$ is a right angle, then $A B \cdot A C = A D \cdot B C$. Therefore, $A B + A C + A B + A C' + C' C > A B + A C$.
$$
$$

This is a contradiction. Hence, $\angle A$ must be an acute angle.
Let $\angle B A D = x_{2}, \angle D A C = x_{1}$, and assume without loss of generality that $x_{2} \geqslant x_{1}$. Then $x = x_{1} + x_{2}$. From $B C + A D - A B - A C = 0$, we get
$$
\operatorname{tg} x_{1} + \operatorname{tg} x_{2} + 1 = \frac{1}{\cos x_{1}} + \frac{1}{\cos x_{2}}.
$$

Simplifying, we obtain
$$
\begin{array}{l}
\sin x + \frac{1}{2} \cos x \\
= 2 \cos \frac{x}{2} \cos \frac{\alpha}{2} \\
- \cos^{2} \frac{x}{2} + \frac{1}{2},
\end{array}
$$

where $\alpha = x_{2} - x_{1} \in [0, x)$.
The right-hand side of (1) is a quadratic function of $\cos \frac{x}{2}$, taking the minimum value $2 \cos \frac{x}{2} - \frac{1}{2}$ when $\alpha = 0$; and the maximum value $\frac{1}{2} + \cos^{2} \frac{x}{2}$ when $\alpha = x$. Therefore,
$$
2 \cos \frac{x}{2} - \frac{1}{2} \leqslant \sin x + \frac{1}{2} \cos x < \frac{1}{2} + \cos^{2} \frac{x}{2}.
$$

Thus, (1) holds, and the solution to this problem is $\left[2 \arcsin \frac{3}{5}, \frac{\pi}{2}\right)$.