In $\triangle ABC$ shown in the figure, $AB=7$, $BC=8$, $CA=9$, and $\overline{AH}$ is an altitude. Points $D$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$, respectively, so that $\overline{BD}$ and $\overline{CE}$ are angle bisectors, intersecting $\overline{AH}$ at $Q$ and $P$, respectively. What is $PQ$?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ \frac{5}{8}\sqrt{3} \qquad \textbf{(C)}\ \frac{4}{5}\sqrt{2} \qquad \textbf{(D)}\ \frac{8}{15}\sqrt{5} \qquad \textbf{(E)}\ \frac{6}{5}$
Get the area of the triangle by [Heron's Formula](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/Heron%27s_Formula): 
\[\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(12)(3)(4)(5)} = 12\sqrt{5}\]
Use the area to find the height $AH$ with known base $BC$:
\[Area = 12\sqrt{5} = \frac{1}{2}bh = \frac{1}{2}(8)(AH)\]
\[AH = 3\sqrt{5}\]
\[BH = \sqrt{AB^2 - AH^2} = \sqrt{7^2 - (3\sqrt{5})^2} = 2\]
\[CH = BC - BH = 8 - 2 = 6\]
Apply the [Angle Bisector Theorem](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/Angle_Bisector_Theorem) on $\triangle ACH$ and $\triangle ABH$, we get $AP:PH = 9:6$ and $AQ:QH = 7:2$, respectively.
To find $AP$, $PH$, $AQ$, and $QH$, apply variables, such that $AP:PH = 9:6$ is $\frac{3\sqrt{5} - x}{x} = \frac{9}{6}$ and $AQ:QH = 7:2$ is $\frac{3\sqrt{5} - y}{y} = \frac{7}{2}$. Solving them out, you will get $AP = \frac{9\sqrt{5}}{5}$, $PH = \frac{6\sqrt{5}}{5}$, $AQ = \frac{7\sqrt{5}}{3}$, and $QH = \frac{2\sqrt{5}}{3}$. Then, since $AP + PQ = AQ$ according to the Segment Addition Postulate, and thus manipulating, you get $PQ = AQ - AP = \frac{7\sqrt{5}}{3} - \frac{9\sqrt{5}}{5}$ = \[\boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}\]