In the convex quadrilateral $ABCD$, point $X$ is selected on side $AD$, and the diagonals intersect at point $E$. It is known that $AC = BD$, $\angle ABX = \angle AX B = 50^o$, $\angle CAD = 51^o$, $\angle AED = 80^o$. Find the value of angle $\angle AXC$.

1. Given that $\angle ABX = \angle AXB = 50^\circ$, we can deduce that $\triangle ABX$ is isosceles with $AB = AX$.
2. Since $\angle ABX = 50^\circ$ and $\angle AXB = 50^\circ$, we can find $\angle BAX$:
   \[
   \angle BAX = 180^\circ - \angle ABX - \angle AXB = 180^\circ - 50^\circ - 50^\circ = 80^\circ
   \]
3. We are given $\angle CAD = 51^\circ$ and $\angle AED = 80^\circ$. We need to find $\angle ADE$:
   \[
   \angle ADE = 180^\circ - \angle EAD - \angle DEA = 180^\circ - 51^\circ - 80^\circ = 49^\circ
   \]
4. Next, we find $\angle BXD$. Since $\angle BAX = 80^\circ$ and $\angle XBA = 50^\circ$:
   \[
   \angle BXD = \angle XBA + \angle BAX = 50^\circ + 80^\circ = 130^\circ
   \]
5. To find $\angle DBX$, we use the fact that the sum of angles in $\triangle BXD$ is $180^\circ$:
   \[
   \angle DBX = 180^\circ - \angle XDB - \angle BXD = 180^\circ - 49^\circ - 130^\circ = 1^\circ
   \]
6. This implies $\angle DBA = \angle DBX + \angle XBA = 1^\circ + 50^\circ = 51^\circ$.
7. Since $\angle DBA = 51^\circ$ and $\angle CAX = 51^\circ$, we can conclude that $\triangle ABD$ and $\triangle ACX$ are congruent by the Angle-Side-Angle (ASA) criterion.
8. Therefore, $\angle AXC = \angle BAD = 80^\circ$.

The final answer is $\boxed{80^\circ}$.