Let's construct a triangle, given the lengths of two sides and the length of the angle bisector of the included angle.
Solution. Consider the problem as solved. Let the sides of triangle $ABC$ be denoted by $a, b, c$ in the usual way, the intersection of the internal angle bisector at vertex $C$ and side $AB$ by $T$, and the intersection of the line through $B$ parallel to $CT$ and line $AC$ by $D$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_5f88568ceeb647cde46eg-1.jpg?height=498&width=659&top_left_y=304&top_left_x=711)

Then $\angle BDC = \angle TCA = \angle TCB = \angle CBD$, so triangle $CBD$ is isosceles. Triangles $ATC$ and $ABD$ are similar because their corresponding sides are parallel, so the ratio of their sides is the same: $\frac{DB}{CT} = \frac{AD}{AC}$. Using this, with $CT = f_c$:

$$
BD = \frac{a+b}{b} \cdot f_c
$$

Using this, the construction can be carried out as follows: Given segments $a, b$, and $f_c$, we construct segment $BD$ using the known method for constructing the fourth proportional (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_5f88568ceeb647cde46eg-1.jpg?height=477&width=848&top_left_y=1220&top_left_x=622)

Then we construct triangle $BCD$, whose three sides we now know, and by measuring the distance $b$ beyond segment $DC$, we obtain point $A$.

The constructed triangle $ABC$ satisfies the conditions because, by the construction, $BC = a$ and $AC = b$, and $\angle ACB = 2 \angle CDB = 2 \angle CBD$, so the internal angle bisector at $C$ is parallel to $BD$, and its length is $CT = BD \cdot \frac{b}{a+b} = f_c$.

The problem has a solution precisely when triangle $BCD$ can be constructed, i.e., if

$$
2a > \frac{a+b}{b} \cdot f_c
$$

rearranging:

$$
\frac{2ab}{a+b} > f_c
$$

Thus, there is a solution if the prescribed length of the angle bisector is smaller than the harmonic mean of the two sides enclosing it, and otherwise, there is no solution.