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, $\frac{C A}{A E} \frac{E D}{D F} \frac{F B}{B C}=1$ 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, $P Q=x+y+10=4 \sqrt{745}$.