One, (50 points) As shown in Figure 2, in the obtuse triangle $\triangle ABC$,
$\angle B - \angle C = 90^{\circ}$,
$\angle C = \theta$, and the circumcircle
$\odot O$ has a radius of $R$.
$AD$ is a diameter of $\odot O$, and a tangent line is drawn through point $D$ intersecting the extension of $BC$ at $H$. A line parallel to $BA$ is drawn through point $D$, intersecting the extension of $AC$ at $E$ and the circle passing through $D$, $O$, and $H$ at $G$. Connect $GH$ and $EH$. Find the area of $\triangle EGH$.
Let the line $O H$ intersect the extensions of $A C$ and $B A$ at $E^{\prime}$ and $F$ respectively. First, we prove: $O$ is the midpoint of segment $E^{\prime} F$.

As shown in Figure 5, let $\angle B A C=\alpha$, $\angle C B A=\beta$. Connect $O B$ and $O C$. Let
$$
\begin{array}{c}
\angle O A B=x, \\
\angle O A C=y, \\
\angle O B C=z. \text{ Then } \\
\alpha+\beta+\theta \\
=180^{\circ}, \\
x-y=\alpha, \\
x+z=\beta, \\
z-y=\theta .
\end{array}
$$

Thus, $x-y+z$
$$
\begin{array}{l}
=\frac{(x-y)+(x+z)+(z-y)}{2} \\
=\frac{\alpha+\beta+\theta}{2}=90^{\circ} .
\end{array}
$$

Therefore, $x=90^{\circ}-\theta, y=\beta-90^{\circ}, z=90^{\circ}-\alpha$.
Connect $B D$, then $\angle D B H=\angle D A C=y$,
$\angle B D H=\angle D A B=x, B D=2 R \sin x$.
In $\triangle D B H$, by the Law of Sines we get
$$
D H=\frac{2 R \sin x \cdot \sin y}{\sin (x+y)} \text {. }
$$

Let $\angle A O F=\gamma$. In the right triangle $\triangle O D H$,
$D H=R \tan \gamma$.
Thus, $\frac{2 R \sin x \cdot \sin y}{\sin (x+y)}=R \tan \gamma$, which means
$2 R \sin x \cdot \sin y \cdot \cos \gamma=R \sin \gamma \cdot \sin (x+y)$
$=R \sin \gamma \cdot \sin x \cdot \cos y+R \sin \gamma \cdot \cos x \cdot \sin y$.
Then $R \sin y \cdot \sin (\gamma-x)=R \sin x \cdot \sin (y-\gamma)$,
which implies $\frac{R \sin x}{\sin (x-\gamma)}=\frac{R \sin y}{\sin (\gamma-y)}$.
In $\triangle A O F$ and $\triangle A O E^{\prime}$, by the Law of Sines we get
$$
O F=\frac{R \sin x}{\sin (x-\gamma)}, O E^{\prime}=\frac{R \sin y}{\sin (\gamma-y)} .
$$

Therefore, $O F=O E^{\prime}$.
Since $O$ is the midpoint of segment $A D$, $D E^{\prime} \parallel B A$. Thus, point $E^{\prime}$ coincides with $E$. Hence,
$\angle G E H=\angle O E D$.
Also, $O, D, G, H$ are concyclic, so,
$\angle H G E=\angle D O H=\angle D O E$.
Therefore, $\triangle E G H \sim \triangle E O D$.
The diameter of the circle passing through $D, O, H$ is $O H=\frac{R}{\cos \gamma}$. And $\angle G D H=90^{\circ}-x$. In $\triangle G D H$, by the Law of Sines we get
$$
\begin{array}{l}
G H=\frac{R}{\cos \gamma} \sin \left(90^{\circ}-x\right)=\frac{R \cos x}{\cos \gamma} . \\
\text { Hence } \frac{S_{\triangle \triangle C H}}{S_{\triangle E O D}}=\left(\frac{G H}{O D}\right)^{2}=\left(\frac{\cos x}{\cos \gamma}\right)^{2}, \text { i.e., } \\
S_{\triangle E C H}=\frac{\cos ^{2} x}{\cos ^{2} \gamma} S_{\triangle E O D} .
\end{array}
$$

In $\triangle A E D$, by the Law of Sines we get
$$
E D=\frac{2 R \sin y}{\sin (x-y)} \text {. }
$$

Then $S_{\triangle E O D}=\frac{1}{2} O D \cdot E D \sin \left(180^{\circ}-x\right)$
$$
=R^{2} \cdot \frac{\sin x \cdot \sin y}{\sin (x-y)} \text {. }
$$

From $\tan \gamma=\frac{2 \sin x \cdot \sin y}{\sin (x+y)}$, we get
$$
\begin{array}{l}
\frac{1}{\cos ^{2} \gamma}=\sec ^{2} \gamma=1+\tan ^{2} \gamma \\
=1+\frac{4 \sin ^{2} x \cdot \sin ^{2} y}{\sin ^{2}(x+y)} \\
=\frac{\sin ^{2}(x+y)+4 \sin ^{2} x \cdot \sin ^{2} y}{\sin ^{2}(x+y)} .
\end{array}
$$

Therefore, $S_{\triangle E C H}$
$$
\begin{array}{l}
=R^{2} \cdot \frac{\cos ^{2} x \cdot \sin ^{2} x \cdot \sin ^{2} y\left[\sin ^{2}(x+y)+4 \sin ^{2} x \cdot \sin ^{2} y\right]}{\sin ^{2}(x+y) \sin ^{2}(x-y)} \\
=R^{2} \cdot \frac{\sin ^{2} \theta \cdot \cos \theta(-\cos \beta)\left[\sin ^{2}(\beta-\theta)+4 \cos ^{2} \theta \cdot \cos ^{2} \beta\right]}{\sin ^{2}(\beta-\theta) \cdot \sin ^{(\beta+\theta)}} \\
=R^{2} \cdot \frac{\sin ^{2} \theta \cdot \cos \theta \cdot \sin \theta\left(1+4 \cos ^{2} \theta \cdot \sin ^{2} \theta\right)}{\cos 2 \theta} \\
=\frac{1}{2} R^{2} \sin ^{2} \theta \cdot \tan 2 \theta\left(1+\sin ^{2} 2 \theta\right) .
\end{array}
$$