4. In an isosceles triangle $ABC$ with base $AC$, the lateral side $AB$ is equal to 2, and segment $AD$ is the angle bisector. Through point $D$, a tangent $DH$ is drawn to the circumcircle of triangle $ADB$, and point $H$ lies on side $AC$. Find the area of triangle $ABC$ if $CD = \sqrt{2} CH$.
Solution: Let $\angle C A D=\angle B A D=\alpha$, then by the properties of the tangent $\angle C D P=\alpha$. Triangles $C D H$ and $A C D$ are similar by two angles. We have $\frac{C H}{C D}=\frac{C D}{A C}=\frac{1}{\sqrt{2}}, C H=\frac{C D}{\sqrt{2}}, A C=C D \sqrt{2}, C H=\frac{A C}{2}$.

Let $C H=A H=x, D H=y$. The angle $\angle A D H=180^{\circ}-4 \alpha$. Applying the Law of Sines to triangles $A D H$ and $H C D$: $\frac{x}{\sin 4 \alpha}=\frac{y}{\sin \alpha}, \quad \frac{x}{\sin \alpha}=\frac{y}{\sin 2 \alpha}$. We arrive at the equation $\sin ^{2} \alpha=\sin 2 \alpha \sin 4 \alpha, 1-\cos 2 \alpha=4 \sin ^{2} 2 \alpha \cos 2 \alpha$, $1-\cos 2 \alpha=4\left(1-\cos ^{2} 2 \alpha\right) \cos 2 \alpha$. Since $2 \alpha<90^{\circ}$, then $\cos 2 \alpha \neq 1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d960854a9c9c598cc600g-21.jpg?height=768&width=470&top_left_y=196&top_left_x=1501)
We have $4 \cos ^{2} 2 \alpha+4 \cos 2 \alpha-1=0, \cos 2 \alpha=\frac{\sqrt{2}-1}{2}$. Then we get

$$
\begin{aligned}
& A H=A B \cos 2 \alpha=\sqrt{2}-1, \quad B H=\sqrt{4-(\sqrt{2}-1)^{2}}=\sqrt{1+2 \sqrt{2}} \\
& S_{A B C}=A H \cdot B H=\sqrt{(1+2 \sqrt{2})(3-2 \sqrt{2})}=\sqrt{4 \sqrt{2}-5}
\end{aligned}
$$

Answer: $\sqrt{4 \sqrt{2}-5}$.