2. On the extension of side $B C$ of an equilateral triangle $A B C$, a point $M$ is chosen. A line parallel to $A C$ is drawn through it. This line intersects the extension of side $A B$ at point $N$. The medians of triangle $B N M$ intersect at point $O$. Point $D$ is the midpoint of $A M$. Find the angles of triangle $O D C$.


Answer: $30^{\circ}, 60^{\circ}, 90^{\circ}$.

Solution. Let's take a point $K$ on $NM$ such that $AK \| CM$, then $ACMK$ is a parallelogram. We will prove that triangles $KON$ and $COM$ are equal by two sides and the angle between them. In triangle $AKN$, angle $N$ is $60^{\circ}$, and $AK \| BC$, so triangle $AKN$ is equilateral, hence $NK = NA = MC$. Point $O$ lies on the axis of the equilateral triangle $NBM$, so $NO = OM$. Medians in an equilateral triangle are also angle bisectors, so $\angle KNO = \angle CMO = 30^{\circ}$.

From the equality of triangles $KON$ and $COM$, it follows that $OK = OC$ and triangle $OKC$ is isosceles. The diagonals of a parallelogram bisect each other at the point of intersection. By the condition, point $D$ is the midpoint of $AM$, so $OD$ is the median of the isosceles triangle $OKC$, and thus it is also an altitude. Therefore, $\angle ODC = 90^{\circ}$. Since $OD$ is the bisector of $\angle KOC$, $\angle DOC = \frac{1}{2} \angle KOC$. But $\angle KOC = \angle KOM + \angle MOC$. Since $\angle MOC = \angle KON$, $\angle KOC = \angle KOM + \angle KON = \angle NOM = 120^{\circ}$, as this is the angle between the medians of an equilateral triangle. Then $\angle DOC = \frac{1}{2} \angle KOC = 60^{\circ}$.

Comment. Correct solution - 20 points. Correct solution with computational errors - 15 points. Special case considered - 5 points. Solution started, with significant progress - 5 points. Solution started, but progress is insignificant - 1 point. Only the correct answer without a solution - 0 points.