One. (50 points) As shown in Figure 1, in $\triangle ABC$, $AB = AC = 3, BC = 2, AD \perp BC$ at point $D, BE \perp AC$ at point $E, EF \perp BC$ at point $F, AD$ intersects $BE$ at point $H$. A circle $\odot O$ is drawn with $HF$ as its diameter, intersecting $HE$ and $EF$ at points $P$ and $Q$ respectively. Extend $QP$ to intersect $AH$ at point $K$. Try to find the lengths of $PQ$ and $HK$.
$$
\begin{array}{c}
B D=D C=D E \\
=\frac{1}{2} B C=1 . \\
\text { Let } \angle C E F=\angle E B C \\
=\angle B E D=\angle B A D= \\
\angle C A D=\alpha \text {. Then } \\
\sin \alpha=\frac{C D}{A C}=\frac{1}{3}, \\
\cos 2 \alpha=1-2 \sin ^{2} \alpha \\
=\frac{7}{9}, \\
D F=D E \sin \angle D E F=1 \times \sin \left(90^{\circ}-2 \alpha\right) \\
=\cos 2 \alpha=\frac{7}{9} .
\end{array}
$$

Since $\tan \alpha=\frac{C D}{A D}=\frac{1}{\sqrt{3^{2}-1^{2}}}=\frac{1}{2 \sqrt{2}}$, we have
$H D=B D \tan \alpha=\frac{1}{2 \sqrt{2}}$,
$$
H F=\sqrt{H D^{2}+D F^{2}}=\sqrt{\frac{1}{8}+\frac{49}{81}}=\frac{\sqrt{946}}{36} \text {. }
$$

Connect $H Q 、 D Q$.
Since $\angle D F Q=90^{\circ}$, we have $\angle D H Q=90^{\circ}$.
Let $\angle H F B=\gamma$, then
$$
\begin{array}{l}
\sin \gamma=\frac{H D}{H F}=\frac{1}{2 \sqrt{2}} \times \frac{36}{\sqrt{946}}=\frac{9}{\sqrt{473}}, \\
\cos \gamma=\sqrt{1-\sin ^{2} \gamma}=\sqrt{\frac{392}{473}}=14 \sqrt{\frac{2}{473}} .
\end{array}
$$

Also, $E Q=E F-Q F=E F-H D$
$$
=E D \sin 2 \alpha-H D=\sqrt{1-\cos ^{2} 2 \alpha}-\frac{1}{2 \sqrt{2}}=\frac{7 \sqrt{2}}{36} \text {, }
$$

In $\triangle E P Q$, by the Law of Sines, we get
$$
\begin{aligned}
P Q= & \frac{\sin \angle P E Q}{\sin \angle E P Q} \cdot E Q=\frac{\sin \left(90^{\circ}-\alpha\right)}{\sin \angle H F Q} \cdot E Q \\
& =\frac{\cos \alpha}{\sin \left(90^{\circ}-\gamma\right)} \cdot E Q=\frac{\cos \alpha}{\cos \gamma} \cdot E Q=\frac{\sqrt{946}}{108} .
\end{aligned}
$$

Clearly, $\angle K P H=\angle H F Q=90^{\circ}-\gamma$,
$\angle K H P=\angle B H D=90^{\circ}-\alpha$.
In $\triangle K H P$, by the Law of Sines, we get
$$
\frac{K P}{K H}=\frac{\sin \left(90^{\circ}-\alpha\right)}{\sin \left(90^{\circ}-\gamma\right)}=\frac{\cos \alpha}{\cos \gamma}=\frac{\sqrt{473}}{21} \text {. }
$$

Thus, $K P=\frac{\sqrt{473}}{21} \cdot K H$.
By the Power of a Point theorem, we have $K P \cdot K Q=K H \cdot K D$, then
$$
K P(K P+P Q)=K H(K H+H D) .
$$

Thus, $K P^{2}+K P \cdot \frac{1}{54} \sqrt{\frac{473}{2}}=K H^{2}+K H \cdot \frac{1}{2 \sqrt{2}}$.
Substituting equation (1) into equation () gives
$$
\begin{array}{l}
\frac{473}{441} K H^{2}+\frac{\sqrt{473}}{21} \times \frac{1}{54} \sqrt{\frac{473}{2}} K H=K H^{2}+\frac{\sqrt{2}}{4} K H, \\
\frac{32}{441} K H^{2}=\frac{\sqrt{2}}{4}\left(1-\frac{473}{567}\right) K H=\frac{47 \sqrt{2}}{2 \times 567} K H . \\
\text { Hence } K H=\frac{441 \times 47 \sqrt{2}}{2 \times 32 \times 567}=\frac{329 \sqrt{2}}{576} .
\end{array}
$$

Thus, $K H=\frac{441 \times 47 \sqrt{2}}{2 \times 32 \times 567}=\frac{329 \sqrt{2}}{576}$.
In summary, we have
$$
P Q=\frac{\sqrt{946}}{108}, K H=\frac{329 \sqrt{2}}{576} \text {. }
$$

Editor's note: This problem can also be solved using similar triangles, the Law of Sines, and the tangent formula.