In a quadrilateral $ABCD$, we have $\angle DAB = 110^{\circ} , \angle ABC = 50^{\circ}$ and $\angle BCD = 70^{\circ}$ . Let $ M, N$ be the mid-points of $AB$ and $CD$ respectively. Suppose $P$ is a point on the segment $M N$ such that $\frac{AM}{CN} = \frac{MP}{PN}$ and $AP = CP$ . Find $\angle AP C$.
1. **Identify the given angles and properties:**
   - $\angle DAB = 110^\circ$
   - $\angle ABC = 50^\circ$
   - $\angle BCD = 70^\circ$
   - $M$ and $N$ are midpoints of $AB$ and $CD$ respectively.
   - $P$ is a point on $MN$ such that $\frac{AM}{CN} = \frac{MP}{PN}$ and $AP = CP$.

2. **Determine the missing angle $\angle CDA$:**
   - Since the sum of the angles in a quadrilateral is $360^\circ$, we can find $\angle CDA$:
     \[
     \angle DAB + \angle ABC + \angle BCD + \angle CDA = 360^\circ
     \]
     \[
     110^\circ + 50^\circ + 70^\circ + \angle CDA = 360^\circ
     \]
     \[
     \angle CDA = 360^\circ - 230^\circ = 130^\circ
     \]

3. **Prove that $ABCD$ is a cyclic quadrilateral:**
   - For a quadrilateral to be cyclic, the opposite angles must sum to $180^\circ$.
   - Check $\angle DAB + \angle BCD$:
     \[
     \angle DAB + \angle BCD = 110^\circ + 70^\circ = 180^\circ
     \]
   - Check $\angle ABC + \angle CDA$:
     \[
     \angle ABC + \angle CDA = 50^\circ + 130^\circ = 180^\circ
     \]
   - Since both pairs of opposite angles sum to $180^\circ$, $ABCD$ is cyclic.

4. **Determine $\angle AEC$:**
   - Since $ABCD$ is cyclic, $\angle AEC$ is the exterior angle of $\triangle AEC$:
     \[
     \angle AEC = \angle DAB - \angle ABC = 110^\circ - 50^\circ = 60^\circ
     \]

5. **Use the given ratio and properties of $P$:**
   - Given $\frac{AM}{CN} = \frac{MP}{PN}$ and $AP = CP$, $P$ is the midpoint of $MN$.
   - Since $M$ and $N$ are midpoints, $MN$ is parallel to $AD$ and $BC$ (midline theorem).

6. **Determine $\angle APC$:**
   - Since $AP = CP$, $\triangle APC$ is isosceles with $AP = CP$.
   - $P$ lies on the perpendicular bisector of $AC$.
   - Since $EAPC$ is cyclic and $AP = CP$, $\angle APC$ is the exterior angle of $\triangle AEC$:
     \[
     \angle APC = 180^\circ - \angle AEC = 180^\circ - 60^\circ = 120^\circ
     \]

The final answer is $\boxed{120^\circ}$