Solution:
Suppose that $P$ lies between $A$ and $B$ and $Q$ lies between $A$ and $C$, and let line $P Q$ intersect lines $A C$ and $B C$ at $E$ and $F$ respectively. As usual, we write $a$, $b$, $c$ for the lengths of $B C$, $C A$, $A B$.

By the angle bisector theorem, $A D / D B = A C / C B$ so that $A D = \frac{b c}{a+b}$ and $B D = \frac{a c}{a+b}$.

Now by Stewart's theorem,
$$c \cdot C D^{2} + \left(\frac{a c}{a+b}\right)\left(\frac{b c}{a+b}\right) c = \frac{a^{2} b c}{a+b} + \frac{a b^{2} c}{a+b}$$
from which
$$C D^{2} = \frac{a b \left((a+b)^{2} - c^{2}\right)}{(a+b)^{2}}$$

Now observe that triangles $C D E$ and $C D F$ are congruent, so $E D = D F$.

By Menelaus' theorem,
$$\frac{C A}{A E} \cdot \frac{E D}{D F} \cdot \frac{F B}{B C} = 1$$
so that
$$\frac{C A}{B C} = \frac{A E}{F B}$$
Since $C F = C E$ while $b > a$, it follows that $A E = \frac{b(b-a)}{a+b}$ so that $E C = \frac{2 a b}{a+b}$.

Finally,
$$D E = \sqrt{C E^{2} - C D^{2}} = \frac{\sqrt{a b \left(c^{2} - (a-b)^{2}\right)}}{a+b}$$

Plugging in $a=20$, $b=80$, $c=65$, we see that $A E = 48$, $E C = 32$, $D E = 10$ as well as $A D = 52$, $B D = 13$.

Now let $P D = x$, $Q E = y$. By power of a point about $D$ and $E$, we have $x(y+10) = 676$ and $y(x+10) = 1536$. Subtracting one from the other, we see that $y = x + 86$. Therefore,
$$x^{2} + 96 x - 676 = 0$$
from which
$$x = -48 + 2 \sqrt{745}$$
Finally,
$$P Q = x + y + 10 = 4 \sqrt{745}$$