Let $ABCD$ be inscribed in a circle with center $O$. Let $E$ be the intersection of $AC$ and $BD$. $M$ and $N$ are the midpoints of the arcs $AB$ and $CD$ respectively (the arcs not containing any other vertices). Let $P$ be the intersection point of $EO$ and $MN$. Suppose $BC=5$, $AC=11$, $BD=12$, and $AD=10$. Find $\frac{MN}{NP}$ 
1. **Given:**
   - \(ABCD\) is a cyclic quadrilateral inscribed in a circle with center \(O\).
   - \(E\) is the intersection of diagonals \(AC\) and \(BD\).
   - \(M\) and \(N\) are the midpoints of the arcs \(AB\) and \(CD\) respectively (the arcs not containing any other vertices).
   - \(P\) is the intersection point of \(EO\) and \(MN\).
   - Given lengths: \(BC = 5\), \(AC = 11\), \(BD = 12\), and \(AD = 10\).

2. **Objective:**
   - Prove that \(\frac{PM}{PN} = \frac{AB}{CD}\).

3. **Key Observations:**
   - Since \(M\) and \(N\) are the midpoints of the arcs \(AB\) and \(CD\) respectively, they are equidistant from \(O\).
   - \(M\) and \(N\) are also the points where the perpendicular bisectors of \(AB\) and \(CD\) intersect the circle.

4. **Properties of Cyclic Quadrilaterals:**
   - The diagonals of a cyclic quadrilateral intersect at an angle such that the opposite angles are supplementary.
   - The power of a point theorem states that for a point \(E\) inside the circle, \(EA \cdot EC = EB \cdot ED\).

5. **Using Similar Triangles:**
   - Consider triangles \(\triangle AEO\) and \(\triangle DEO\). Since \(O\) is the center, \(OA = OC\) and \(OB = OD\).
   - By the intersecting chords theorem, \(\frac{AE}{DE} = \frac{AB}{CD}\).

6. **Using the Intersecting Chords Theorem:**
   - Since \(E\) is the intersection of \(AC\) and \(BD\), we have:
     \[
     AE \cdot EC = BE \cdot ED
     \]
   - This implies that the triangles \(\triangle AEO\) and \(\triangle DEO\) are similar by AA similarity criterion.

7. **Relating \(PM\) and \(PN\):**
   - Since \(P\) lies on \(EO\), and \(M\) and \(N\) are on the circle, the line \(MN\) is perpendicular to \(EO\) at \(P\).
   - By the properties of similar triangles and the fact that \(M\) and \(N\) are midpoints of the arcs, we have:
     \[
     \frac{PM}{PN} = \frac{AE}{DE}
     \]
   - From the intersecting chords theorem, we know:
     \[
     \frac{AE}{DE} = \frac{AB}{CD}
     \]

8. **Conclusion:**
   - Therefore, we have:
     \[
     \frac{PM}{PN} = \frac{AB}{CD}
     \]

\(\blacksquare\)

The final answer is \( \boxed{ \frac{PM}{PN} = \frac{AB}{CD} } \)