A circle with center $O$ has a diameter $A B$. A point $C$ on the circumference of the circle, different from $A$ and $B$, draws a perpendicular to $A B$, intersecting $A B$ at $D$. A perpendicular from $O$ to $B C$ intersects $B C$ at $M$. Determine the measure of angle $A B C$, given that $D B = 3 \cdot O M$.
I. solution. The $ABC$ triangle is right-angled, with the right angle at vertex $C$, that is, $AC \perp BC$. This, along with $OM \perp BC$, implies $OM \| AC$, and since $O$ bisects $AB$, $OM$ is the midline of the $ABC$ triangle. Thus, $AC = 2OM$.

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For the leg $AC$ of the right-angled $ABC$ triangle, we can write the well-known leg theorem. With the notation $AD = x$, using the previous relationships,

$$
2 \cdot OM = AC = \sqrt{AD \cdot AB} = \sqrt{AD \cdot (AD + DB)} = \sqrt{x(x + 3 \cdot OM)}
$$

After rearranging, we get the quadratic equation

$$
x^2 + 3 \cdot OM \cdot x - 4 \cdot OM^2 = 0
$$

The roots of the equation are $x_1 = OM, x_2 = -4 \cdot OM$, of which only the positive value is relevant. In the $ABC$ triangle, therefore, $AB = AD + DB = 4OM$, hence $AB = 2 \cdot AC$. Thus, the $ABC$ triangle is a right-angled triangle where the hypotenuse is twice the length of the leg opposite vertex $B$, from which it follows that $\angle ABC = 30^\circ$.

II. solution. Fix the diameter $AB$ of the circle. Start the point $C$ from $A$ and let $C$ run along the circumference of the circle all the way to $B$. As $C$ moves, it gets farther from $A$ - that is, the distance $AC = 2 \cdot OM$ increases - while the projection of $C$ onto $AB$, $D$, gets closer to $B$ - that is, $DB$ decreases continuously. Therefore, the ratio $OM / DB$ has a numerator that increases and a denominator that decreases, so the value of the fraction increases continuously as $C$ moves. Thus, there can be at most one position of $C$ for which the ratio $OM / DB$ is $1/3$.

It is easy to see that if $\angle ABC = 30^\circ$, then $OM / DB = 1/3$. Indeed, $OM / DB = (OM / MB) \cdot (MB / DB)$ and when $\angle OBM = \angle CBD = 30^\circ$,

$$
\frac{OM}{MB} = \frac{1}{\sqrt{3}}, \quad \frac{MB}{DB} = \frac{1}{2} \cdot \frac{CB}{DB} = \frac{1}{2} \cdot \frac{2}{\sqrt{3}}
$$

The value of the ratio $OM / DB$ cannot be $1/3$ in any other position of $C$, so when $DB = 3 \cdot OM$, $\angle ABC = 30^\circ$, which answers the question of the problem.