In triangle $ABC$, $AC < BC$ and $D$ is the point on side $BC$ such that $\angle CAD = \angle CBA$, and $C_{0}$ is the midpoint of side $AB$. In which triangles will the line $DC_{0}$ be parallel to the internal angle bisector of $\angle C$?
Let the angles at the vertices $A, B, C$ of the triangle be denoted by $\alpha, \beta, \gamma$, respectively. Since $A C < B C$, we have $\beta < \alpha$. This condition ensures that $\beta$ "fits" into $\alpha$, so point $D$ can be placed on side $B C$ such that $\angle C A D = \beta$.

![](https://cdn.mathpix.com/cropped/2024_05_02_34a35f2339f622b165f1g-1.jpg?height=371&width=634&top_left_y=259&top_left_x=732)

Let's calculate the angles at side $C_{0} D$ of triangle $A C_{0} D$:

\[
\begin{gathered}
\angle A C_{0} D = \angle C_{0} B D + \angle C_{0} D B = \beta + \angle C_{0} D B \\
\angle A D C_{0} = \angle D A C + \angle A C D - \angle C_{0} D B = \beta + \gamma - \angle C_{0} D B .
\end{gathered}
\]

If $C_{0} D$ is parallel to the internal angle bisector of $\angle C$, then $\angle C_{0} D B = \gamma / 2$, so both angles are equal to $\beta + \gamma / 2$. Conversely, if these two angles are equal, then $\angle C_{0} D B = \gamma / 2$, and $C_{0} D$ is parallel to the angle bisector. Therefore, we need to determine when these two angles are equal, or when $A D = A C_{0}$.

Triangle $A D C$ is similar to triangle $A B C$ because they have two equal angles. Therefore, $A D : A B = A C : B C$, and $A D = A C_{0} = \frac{1}{2} A B$ holds exactly when $A C = \frac{1}{2} B C$.