Let $ABCD$ be a convex quadrilateral with $\angle A=60^o$. Let $E$ and $Z$ be the symmetric points of $A$ wrt $BC$ and $CD$ respectively. If the points $B,D,E$ and $Z$ are collinear, then calculate the angle $\angle BCD$.
1. Given that $ABCD$ is a convex quadrilateral with $\angle A = 60^\circ$. Let $E$ and $Z$ be the symmetric points of $A$ with respect to $BC$ and $CD$ respectively. This means that $E$ and $Z$ are reflections of $A$ over lines $BC$ and $CD$ respectively.
2. Since $E$ is the reflection of $A$ over $BC$, $\triangle ABE$ is an isosceles triangle with $\angle ABE = \angle AEB = 60^\circ$.
3. Similarly, since $Z$ is the reflection of $A$ over $CD$, $\triangle ADZ$ is an isosceles triangle with $\angle ADZ = \angle AZD = 60^\circ$.
4. Given that points $B, D, E, Z$ are collinear, we need to find the angle $\angle BCD$.
5. Let $\angle ABD = 2X^\circ$ and $\angle ADB = 2Y^\circ$. Since $\triangle ABE$ and $\triangle ADZ$ are isosceles triangles with $\angle A = 60^\circ$, we have:
   \[
   \angle ABE = \angle AEB = 60^\circ
   \]
   \[
   \angle ADZ = \angle AZD = 60^\circ
   \]
6. Since $B, D, E, Z$ are collinear, the angles $\angle CBD$ and $\angle CDB$ can be expressed as:
   \[
   \angle CBD = 90^\circ - X^\circ
   \]
   \[
   \angle CDB = 90^\circ - Y^\circ
   \]
7. The sum of the angles in $\triangle BCD$ is $180^\circ$, so:
   \[
   \angle BCD = 180^\circ - \angle CBD - \angle CDB
   \]
   Substituting the values:
   \[
   \angle BCD = 180^\circ - (90^\circ - X^\circ) - (90^\circ - Y^\circ)
   \]
   Simplifying:
   \[
   \angle BCD = 180^\circ - 90^\circ + X^\circ - 90^\circ + Y^\circ
   \]
   \[
   \angle BCD = X^\circ + Y^\circ
   \]
8. Since $\angle A = 60^\circ$ and $\angle ABD = 2X^\circ$ and $\angle ADB = 2Y^\circ$, we have:
   \[
   2X^\circ + 2Y^\circ + 60^\circ = 180^\circ
   \]
   Simplifying:
   \[
   2X^\circ + 2Y^\circ = 120^\circ
   \]
   \[
   X^\circ + Y^\circ = 60^\circ
   \]
9. Therefore, $\angle BCD = X^\circ + Y^\circ = 60^\circ$.

The final answer is $\boxed{60^\circ}$