Let $ABC$ be a triangle where $AC\neq BC$. Let $P$ be the foot of the altitude taken from $C$ to $AB$; and let $V$ be the orthocentre, $O$ the circumcentre of $ABC$, and $D$ the point of intersection between the radius $OC$ and the side $AB$. The midpoint of $CD$ is $E$.

a) Prove that the reflection $V'$ of $V$ in $AB$ is on the circumcircle of the triangle $ABC$.

b) In what ratio does the segment $EP$ divide the segment $OV$?
### Part A:
1. **Orthocenter Reflection Lemma**: The Orthocenter Reflection Lemma states that the reflection of the orthocenter \( V \) of a triangle \( \triangle ABC \) over any side of the triangle lies on the circumcircle of the triangle. 

2. **Application of the Lemma**: In our problem, \( V \) is the orthocenter of \( \triangle ABC \). We need to show that the reflection \( V' \) of \( V \) over the side \( AB \) lies on the circumcircle of \( \triangle ABC \).

3. **Reflection**: Reflect \( V \) over \( AB \) to get \( V' \). By the Orthocenter Reflection Lemma, \( V' \) lies on the circumcircle of \( \triangle ABC \).

Thus, we have proved that the reflection \( V' \) of \( V \) in \( AB \) is on the circumcircle of the triangle \( ABC \).

\(\blacksquare\)

### Part B:
1. **Definitions and Known Points**:
   - Let \( C_1 \) be the \( C \)-antipode with respect to the circumcircle of \( \triangle ABC \).
   - Let \( X \) be the midpoint of \( AB \).
   - Let \( Y \) be the midpoint of \( CV \).

2. **Symmetry and Midlines**:
   - It is well-known that \( C_1 \) and \( V \) are symmetric about \( X \). This implies that \( X \) is the midpoint of \( C_1V \).
   - Since \( X \) is the midpoint of \( AB \) and \( Y \) is the midpoint of \( CV \), the line segment \( XY \) is parallel to \( CC_1 \).

3. **Homothety and Nine-Point Center**:
   - Consider a homothety centered at \( P \) that maps \( C \) to \( D \). This homothety implies that \( PE \) bisects \( XY \).
   - The point where \( PE \) intersects \( XY \) is the Nine-Point center \( N_9 \) of \( \triangle ABC \).

4. **Midpoint of \( OV \)**:
   - It is well-known that the Nine-Point center \( N_9 \) is the midpoint of the segment \( OV \).

5. **Conclusion**:
   - Since \( PE \) intersects \( OV \) at \( N_9 \), and \( N_9 \) is the midpoint of \( OV \), the segment \( EP \) divides the segment \( OV \) in the ratio \( 1:1 \).

Thus, the ratio in which the segment \( EP \) divides the segment \( OV \) is \( 1:1 \).

\(\blacksquare\)

The final answer is \( \boxed{ 1:1 } \).