Problem 3. In a right triangle $ACB$ (where $\angle C$ is the right angle), the bisector $AD=\frac{C \sqrt{3}}{3}$, and the hypotenuse $AB=c$. Find the legs $AC$ and $CB$ of this triangle.
First solution. Let: $AC = x, BC = y$, $CD = z$. We obtain the following system of equations:

$$
\left\{\begin{aligned}
x^{2} + z^{2} & = c^{2} \\
x^{2} + z^{2} & = \frac{c^{2}}{3} \\
\frac{y - z}{z} & = \frac{c}{x}
\end{aligned}\right.
$$

From equation (3) we find:

$$
z = \frac{yx}{c + x}
$$

From equations (2) and (4) it follows:

$$
x^{2} + \frac{y^{2} x^{2}}{(c + x)^{2}} = \frac{c^{2}}{3}
$$

From equations (1) and (5) we get:

$$
x^{2} + \frac{x^{2}(c^{2} - x^{2})}{(c + x)^{2}} = \frac{c^{2}}{3}
$$

After simplifications, equation (6) takes the form:

$$
6 x^{2} - c x - 2 c^{2} = 0
$$

Solving equation (7), we get $x = 0.5 c$.

Thus, $AC = 0.5 c$ and $BC = 0.5 c \sqrt{3}$.

Second solution. Let: $AC = x, BC = y$, $\angle CAD = \angle BAD = t$. We obtain the system of equations:

$$
\left\{\begin{array}{l}
x^{2} + y^{2} = c^{2} \\
x = \frac{c \sqrt{3}}{3} \cos t \\
y = c \sin 2t
\end{array}\right.
$$

By substituting the values of $x$ and $y$ into the equation, we get:

$$
\begin{gathered}
\frac{c^{2}}{3} \cos^{2} t + c^{2} \sin^{2} 2t = c^{2}, \cos^{2} t + 3 \sin^{2} 2t = 3 \\
\frac{1 + \cos 2t}{2} + 3(1 - \cos^{2} 2t) = 3 \\
6 \cos^{2} 2t - \cos 2t + 1 = 0
\end{gathered}
$$

Thus,

$$
\cos 2t_{1} = -\frac{1}{3} ; \cos 2t_{2} = 0.5
$$

By the condition of the problem, $0^{\circ} < 2t < 90^{\circ}$, i.e., $0 < \cos 2t < 1$. From the equation $\cos 2t = 0.5$ we get $t = 30^{\circ}$. From equations (9) and (10) we find $x$ and $y$.

Third solution. Let: $\angle CAD = \angle DAB = x$. Then $AC = \frac{c \sqrt{3}}{3} \cos x$. The equality $S_{\triangle ACD} + S_{\triangle ADB} = S_{\triangle ACB}$ is obvious. Therefore, applying the formula for the area of a triangle $S = 0.5 ab \sin C$, we get:

$$
\begin{gathered}
0.5 \frac{c \sqrt{3}}{3} \cos x \cdot \frac{c \sqrt{3}}{3} \sin x + 0.5 \frac{c \sqrt{3}}{3} c \cdot \sin x = \\
= 0.5 \frac{c \sqrt{3}}{3} \cos x \cdot c \cdot \sin 2x
\end{gathered}
$$

After simplification, equation (11) transforms to:

$$
\frac{\sqrt{3}}{3} \cos x + 1 = 2 \cos^{2} x
$$

Solving equation (13), we find $\cos x = 0.5 \sqrt{3}$, i.e., $x = 30^{\circ}$.

Thus, $\angle CAB = 60^{\circ}, \quad \angle CBA = 30^{\circ}$ and $AC = 0.5 c$; $BC = 0.5 \sqrt{3}$.

Fourth solution. If $\angle DAB = x$, then $\angle ABC = 90^{\circ} - 2x, \quad \angle ADB = 180^{\circ} - x - (90^{\circ} - 2x) = 90^{\circ} + x$. By the Law of Sines in triangle $ADB$ we get:

$$
\frac{\sin (90^{\circ} + x)}{c} = \frac{\sin (90^{\circ} - 2x)}{\frac{c \sqrt{3}}{3}}
$$

From this, $\cos x = \sqrt{3} \cos 2x, \cos x = \sqrt{3}(2 \cos^{2} x - 1)$. Solving this equation, we get $\cos x = 0.5 \sqrt{3}$ and $x = 30^{\circ}$.