A side of a triangle is divided into segments $d, e, f$ by the trisectors of the opposite angle. Calculate the angle opposite this side. For our numerical examples ($d, e, f$ respectively):
I. $1 ; 1 ; 2 ;$
II. $11 ; 8 ; 11 ;$
III. $4 ; 3 ; 5$.
Let the vertices of the triangle be denoted by $A, B, C$, and let $AB = d + e + f$ such that its segments are $AD = d, DE = e, EB = f$. Furthermore, let $CA = b, CD = g, CE = h, CB = a$, and $\angle ACB = \gamma$. According to the assumption, $\angle ACD = \angle DCE = \angle ECB = \frac{\gamma}{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_1dbb807ccc8b44a0d50ag-1.jpg?height=271&width=585&top_left_y=323&top_left_x=759)

Express the area of triangle $ACE$ in terms of sides $CA, CE$ and angle $ACE$, and also as the sum of the areas of triangles $ACD$ and $DCE$ expressed similarly:

$$
\frac{1}{2} b h \sin \frac{2 \gamma}{3} = b h \sin \frac{\gamma}{3} \cos \frac{\gamma}{3} = \frac{1}{2}(b g + g h) \sin \frac{\gamma}{3}
$$

From this, since $\gamma > 0^\circ$, $\sin (\gamma / 3) \neq 0$,

$$
\cos \frac{\gamma}{3} = \frac{g(b + h)}{2 b h}
$$

We re-express our result for triangle $DCB$ and then multiply the new expression by (1):

$$
\begin{gathered}
\cos \frac{\gamma}{3} = \frac{h(g + a)}{2 g a} \\
\cos^2 \frac{\gamma}{3} = \frac{(b + h)(g + a)}{4 b a} = \frac{1}{4}\left(1 + \frac{h}{b}\right)\left(\frac{g}{a} + 1\right) = \\
= \frac{1}{4}\left(1 + \frac{e}{d}\right)\left(\frac{e}{f} + 1\right) = \frac{(d + e)(e + f)}{4 d f}
\end{gathered}
$$

In the transformation, we used the fact that $CD$ bisects angle $ACE$ and $CE$ bisects angle $DCB$, so by a known theorem,

$$
\frac{CE}{CA} = \frac{DE}{DA} \quad \text{and} \quad \frac{CD}{CB} = \frac{ED}{EB}
$$

Thus, we expressed $\gamma$ in terms of segments $d, e, f$.

In the numerical examples, we get respectively

$$
\cos^2 \frac{\gamma}{3} = \frac{3}{4}, \quad \frac{19^2}{4 \cdot 11^2}, \quad 0.7
$$

Since $\gamma / 3$ is always an acute angle, its cosine is positive, so we get respectively

$$
\begin{aligned}
& \cos \frac{\gamma}{3} = \frac{\sqrt{3}}{2}, & \frac{19}{22}, & 0.8367 \\
\cos \gamma = & \left(4 \cos^2 \frac{\gamma}{3} - 3\right) \cos \frac{\gamma}{3} = 0, & -\frac{19}{11^3}, & -0.1673 \\
\gamma = & 90^\circ, & 90^\circ 49', & 99^\circ 38'
\end{aligned}
$$

Remarks. In (1), we essentially determined the half-angle of the angle between sides $b, h$ and the angle bisector $g$, which is an interesting result in itself.

2. We can obtain (2) from the fact that the area of triangle $ABC$ is equal to the sum of the areas of triangles $ADC, DEC$, and $EBC$, if we use the identity expressing $\sin \gamma$ in terms of $\sin (\gamma / 3)$:

$$
\begin{gathered}
(2t =) a b \sin \gamma = a b \left(3 - 4 \sin^2 \frac{\gamma}{3}\right) \sin \frac{\gamma}{3} = (b g + g h + h a) \sin \frac{\gamma}{3} \\
\sin^2 \frac{\gamma}{3} = \frac{3}{4} - \frac{g(b + h) + a(h + b - b)}{4 a b} = 1 - \frac{(a + g)(b + h)}{4 a b}
\end{gathered}
$$

3. In the first and second numerical examples, we can reach the goal more simply based on the equality of two segments. In the first case, $d : e = b : h = 1$, so triangle $AEC$ is isosceles, and $\angle DCB = 90^\circ$, and $e : f = g : a = 1 : 2$, which is the value of $\cos \angle DCB$, so $\angle DCB = 60^\circ$.

In the second case, we can use the equality $CA = CB$, which follows from the fact that $C$ is the intersection of two Apollonian circles: $CA : CE = d : e = f : e = CB : CD$, and these are reflections of each other across the perpendicular bisector of $AB$ or, in this case, $DE$.