A triangle's angle bisector starting from $A$ is parallel to the line $O M$, but not coincident with it, where $O$ is the center of the circumscribed circle and $M$ is the orthocenter. What is the measure of the angle at $A$?
Let the other two vertices of the triangle be denoted by $B$ and $C$. The angle bisector from $A$ intersects the circumcircle at $D$. Since $C D = D B$, $D O$ is perpendicular to $C B$ and bisects it. Let the midpoint of the bisection be $F$. By the given condition, $M O \| A D$, and $M A$ is parallel to $O D$ since both are perpendicular to $C B$, thus $M O D A$ is a parallelogram, and hence $M A = O D$. Let the reflection of $C$ over $O$ be denoted by $C^{*}$. The triangle $C B C^{*}$ is clearly a right triangle and similar to the triangle $C F O$. Since $B C^{*} \| F O$ and $C B = 2 C F$, it follows that $B C^{*} = 2 O F$.

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The quadrilateral $M A C^{*} B$ is also a parallelogram, with opposite sides parallel: $M A \| F D \| B C^{*}$, and since $C A \perp A C^{*}$ (as $C C^{*}$ is a diameter of the circle), and $M B$ is perpendicular to $A C$, it follows that $M B \| A C^{*}$. Therefore, $M A = O D = B C^{*}$, and since $O D = 2 O F$, $O$ is the centroid of the triangle $B C D$. The triangle $C B D$ is clearly isosceles. From $O B = O D$, it follows that the median $C O$ not only bisects the side $D B$ but is also perpendicular to it. Thus, the triangle $C B D$ is also equilateral, and since $\angle C D B = 60^{\circ}$, and $A$ lies on the other arc of the chord $C B$ (since $A D$ always passes through the interior of the triangle), it follows that $\angle C A B = 120^{\circ}$.