\left.\begin{array}{l}{[\quad \text { Pythagorean Theorem (direct and inverse). }} \\ \text { [ The ratio in which the bisector divides the side. }]\end{array}\right]

The height of a triangle, equal to 2, divides the angle of the triangle in the ratio 2:1, and the base of the triangle - into parts, the smaller of which is equal to 1. Find the area of the triangle.
## First Method.

Let $BD=2$ be the height of triangle $ABC$, and $AD=1$. In the right triangles $ABD$ and $CBD$, the common leg is $BD$, and according to the problem, the leg $AD$ of the first triangle is less than the leg $CD$ of the second. Therefore, $\angle ABD < \angle CBD$.

Let $\angle ABD = \alpha$. Then $\angle CBD = 2\alpha$. From the right triangle $ABD$, we find that

$$
\tan \alpha = \frac{AD}{BD} = \frac{1}{2}
$$

Then

$$
\tan 2\alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} = \frac{2 \cdot \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} = \frac{4}{3}
$$

Then, in the right triangle $CBD$,

$$
CD = BD \cdot \tan \angle CBD = 2 \cdot \tan 2\alpha = 2 \cdot \frac{4}{3} = \frac{8}{3}
$$

Therefore,

$$
S_{\triangle ABC} = \frac{1}{2} AC \cdot BD = \frac{1}{2} \left(1 + \frac{8}{3}\right) \cdot 2 = \frac{11}{3}
$$

## Second Method.

Let $BD=2$ be the height of triangle $ABC$, and $AD=1$. In the right triangles $ABD$ and $CBD$, the common leg is $BD$, and according to the problem, the leg $AD$ of the first triangle is less than the leg $CD$ of the second. Therefore, $\angle ABD < \angle CBD$. Let $\angle ABD = \alpha$. Then $\angle CBD = 2\alpha$. Suppose the angle bisector of $\angle DBC$ intersects segment $CD$ at point $E$. Then triangle $ABE$ is isosceles, because its height $BD$ is also the angle bisector. Therefore, $DE = AD = 1$. Since $BD = 2DE$, by the property of the angle bisector of a triangle, $BC = 2CE$. Let $CE = x$. By the Pythagorean theorem,

$$
BC^2 = BD^2 + CD^2, \text{ or } 4x^2 = 4 + (1 + x)^2
$$

From this equation, we find that $x = \frac{5}{3}$. Therefore,

$$
S_{\triangle ABC} = \frac{1}{2} AC \cdot BD = \frac{1}{2} \left(1 + 1 + \frac{5}{3}\right) \cdot 2 = \frac{11}{3}
$$

## Answer

$\frac{11}{3}$