Let $ABC$ be an acute-angled triangle with circumcenter $O$ and let $M$ be a point on $AB$. The circumcircle of $AMO$ intersects $AC$ a second time on $K$ and the circumcircle of $BOM$ intersects $BC$ a second time on $N$.

Prove that $\left[MNK\right] \geq \frac{\left[ABC\right]}{4}$ and determine the equality case.
1. **Identify the midpoints and the medial triangle:**
   Let $X$, $Y$, and $Z$ be the midpoints of $BC$, $AC$, and $AB$ respectively. The triangle $\triangle XYZ$ is the medial triangle of $\triangle ABC$.

2. **Properties of the medial triangle:**
   The medial triangle $\triangle XYZ$ is similar to $\triangle ABC$ and its area is $\frac{1}{4}$ of the area of $\triangle ABC$. This is because each side of the medial triangle is half the length of the corresponding side of $\triangle ABC$.

3. **Circumcircle intersections:**
   The circumcircle of $\triangle AMO$ intersects $AC$ at $K$, and the circumcircle of $\triangle BOM$ intersects $BC$ at $N$. We need to show that the area of $\triangle MNK$ is at least $\frac{1}{4}$ of the area of $\triangle ABC$.

4. **Similarity and area comparison:**
   Since $O$ is the circumcenter of $\triangle ABC$, $O$ lies on the perpendicular bisectors of $AB$, $BC$, and $CA$. The points $M$, $N$, and $K$ are such that $\triangle MNK$ is similar to $\triangle XYZ$ because they are both formed by intersections involving the circumcircle and sides of $\triangle ABC$.

5. **Area relationship:**
   Since $\triangle XYZ$ is similar to $\triangle MNK$ and $\triangle XYZ$ has an area of $\frac{1}{4}$ of $\triangle ABC$, we have:
   \[
   [XYZ] = \frac{1}{4} [ABC]
   \]
   Given that $\triangle MNK$ is similar to $\triangle XYZ$, we have:
   \[
   [MNK] \geq [XYZ]
   \]
   Therefore:
   \[
   [MNK] \geq \frac{1}{4} [ABC]
   \]

6. **Equality case:**
   The equality holds when $\triangle MNK$ is exactly the medial triangle $\triangle XYZ$. This happens when $M$ is the midpoint of $AB$, $N$ is the midpoint of $BC$, and $K$ is the midpoint of $AC$.

\[
\boxed{\left[MNK\right] \geq \frac{\left[ABC\right]}{4}}
\]