Let's determine the area of a triangle given two sides and the angle bisector of the included angle.
Let the sides of triangle $ABC$ be denoted as usual by $a, b, c$! Let the intersection of the angle bisector from vertex $C$ and side $AB$ be $F$, and the length of segment $FC$ be $f$. Let $\angle BCF = \angle FCA = \gamma$. We need to determine the area $T$ of the triangle as a function of $a, b$, and $f$.

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It is known that twice the area of any triangle is equal to the product of two sides and the sine of the included angle. Since the area of triangle $ABC$ is the sum of the areas of triangles $ACF$ and $FCB$, we have

$$
a \cdot b \cdot \sin 2 \gamma = b \cdot f \cdot \sin \gamma + a \cdot f \cdot \sin \gamma = (a+b) \cdot f \cdot \sin \gamma
$$

By the identity $\sin 2 \gamma = 2 \sin \gamma \cdot \cos \gamma$, we get

$$
2 \cdot a \cdot b \cdot \sin \gamma \cdot \cos \gamma = (a+b) \cdot f \cdot \sin \gamma
$$

Since $2 \gamma$ is an angle in a triangle, $\sin \gamma > 0$, so we can simplify by dividing by $\sin \gamma$.

$$
2 \cdot a \cdot b \cdot \cos \gamma = (a+b) \cdot f, \quad \text { that is } \quad \cos \gamma = \frac{(a+b) \cdot f}{2 a b}
$$

Since $\sin \gamma > 0$, we have $\sin \gamma = \sqrt{1 - \cos^2 \gamma}$. Substituting the expression for $\cos \gamma$ we obtained, we get

$$
\sin \gamma = \sqrt{1 - \frac{(a+b)^2 f^2}{4 a^2 b^2}}
$$

Thus, the area of triangle $ABC$ is

$$
\begin{aligned}
& T = \frac{a \cdot b \cdot \sin 2 \gamma}{2} = \frac{2 \cdot a \cdot b \cdot \sin \gamma \cos \gamma}{2} = \\
& = \frac{(a+b) f}{4 a b} \sqrt{4 a^2 b^2 - (a+b)^2 f^2} .
\end{aligned}
$$

This formula expresses the area of the triangle with the desired data.