6. Given that the incircle of $\triangle A B C$ touches sides $A B$ and $A C$ at points $E$ and $F$, respectively, and $A D$ is the altitude from $A$ to side $B C$, and $A E + A F = A D$. Then the range of $\sin \frac{A}{2}$ is $\qquad$
6. $\left[\frac{3}{5}, \frac{\sqrt{2}}{2}\right)$.

Let $B C=a, A C=b, A B=c, A D=h$.
From the given, we have $b+c=a+h$.
By the area formula of a triangle, we get $\sin A=\frac{a h}{b c}$.
By the cosine rule, we have
$$
\begin{array}{l}
\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{(b+c)^{2}-a^{2}-2 b c}{2 b c} \\
=\frac{h^{2}+2 a h}{2 b c}-1 \\
\Rightarrow \cot \frac{A}{2}=\frac{1+\cos A}{\sin A}=1+\frac{h}{2 a} .
\end{array}
$$

Construct $C E \perp B C$, such that points $E$ and $A$ are on the same side of $B C$, and $E C=2 h$, connect $B E$ and $A E$.
Thus, $A E=A C=b$.
In the right triangle $\triangle B C E$, $B E=\sqrt{a^{2}+4 h^{2}}$.
Then $a+h=A B+A E \geqslant B E=\sqrt{a^{2}+4 h^{2}}$
$\Rightarrow 0<\frac{h}{a} \leqslant \frac{2}{3}$
$\Rightarrow \cot \frac{A}{2} \in\left(1, \frac{4}{3}\right]$
$\Rightarrow \sin \frac{A}{2}=\frac{1}{\sqrt{1+\cot ^{2} \frac{A}{2}}} \in\left[\frac{3}{5}, \frac{\sqrt{2}}{2}\right)$.