Solution:

Let $Q$ be the midpoint of line segment $BP$. The conditions of the problem imply $|BQ| = |QP| = |PC| = \frac{1}{3}|BC|$. Let $R$ be the midpoint of line segment $AB$. Then $RQ$ is a midline of $ABP$. Consequently, $RQ \parallel AP$. Ray $AP$ bisects side $CQ$ of triangle $CRQ$ while being parallel to side $RQ$ of this triangle. Thus $AP$ extends the midline of triangle $CRQ$ and bisects therefore also its side $CR$. But line segment $CR$ is the median of triangle $ABC$

drawn from vertex $C$.

Solution:

Let $R$ be the midpoint of segment $AB$. Choose point $D$ on ray $AC$ beyond point $C$ such that $|AC| = |CD|$. Then $BC$ is a median of triangle $ABD$. As $|BP| : |PC| = 2 : 1$, point $P$ is the intersection point of medians of triangle $ABD$. Thus $AP$ lies entirely on the other median of triangle $ABD$, i.e., ray $AP$ bisects the segment $BD$. As $CR$ is the midline of triangle $ABD$, we have $CR \parallel BD$, implying that ray $AP$ also bisects the segment $CR$. But this is the median of triangle $ABC$ drawn from vertex $C$.