6. In $\triangle A B C$, the lengths of the three sides $B C=a, C A=b$, $A B=c$ satisfy $c<b<a<2 c . P 、 Q$ are two points on the sides of $\triangle A B C$, and line $P Q$ divides $\triangle A B C$ into two parts with equal area. Find the minimum length of segment $P Q$.
6. First consider the case where points $P, Q$ are on $AB, AC$ respectively.

Assume point $P$ is on $AB$, and point $Q$ is on $AC$.
Let $AP = x, AQ = y$.
By $S_{\triangle APQ} = \frac{1}{2} S_{\triangle ABC}$
$\Rightarrow xy \sin A = \frac{1}{2} bc \sin A \Rightarrow xy = \frac{1}{2} bc$.
By the cosine rule,
$|PQ|^2 = x^2 + y^2 - 2xy \cos A$
$\geqslant 2xy - 2xy \cos A$
$= bc(1 - \cos A) = 2bc \sin^2 \frac{A}{2}$.
Given $c < b < a < 2c$, we know $\sqrt{\frac{bc}{2}} < c < b$.
Thus, when $x = y = \sqrt{\frac{bc}{2}}$, points $P, Q$ are inside sides $AB, AC$ respectively, and at this time, $PQ^2$ reaches its minimum value
$d(a) = 2bc \sin^2 \frac{A}{2}$.
Let $R$ be the circumradius of $\triangle ABC$.
Since $a = 2R \sin A$, we have
$d(a) = 2bc \sin^2 \frac{A}{2} = \frac{abc}{R \sin A} \sin^2 \frac{A}{2}$
$= \frac{abc}{2R} \tan \frac{A}{2}$.
Given $c < b < a < 2c$, we get
$\sqrt{\frac{ca}{2}} < c < a, \sqrt{\frac{ab}{2}} < b < a$.
Similarly, when line $l$ intersects sides $AB, BC$, the minimum value of $PQ^2$ is
$$
d(b) = \frac{abc}{2R} \tan \frac{B}{2} ;
$$

When line $l$ intersects sides $AC, BC$, the minimum value of $PQ^2$ is
$$
d(c) = \frac{abc}{2R} \tan \frac{C}{2} .
$$

Since the tangent function is monotonically increasing in the interval $\left(0, \frac{\pi}{2}\right)$, and $0 < \angle C < \angle B < \angle A < \pi$, we have
$$
d(c) < d(b) < d(a) .
$$

Therefore, the minimum length of segment $PQ$ is
$$
\sqrt{d(c)} = \sqrt{2ab} \sin \frac{C}{2} \text{. }
$$

[Note] The answer to this problem can also be expressed as
$$
\sqrt{\frac{c^2 - (a - b)^2}{2}} \sqrt{\frac{1}{2}(a + c - b)(b + c - a)}
$$

or other forms.