Quadrilateral $ALEX,$ pictured below (but not necessarily to scale!)
can be inscribed in a circle; with $\angle LAX = 20^{\circ}$ and $\angle AXE = 100^{\circ}:$
1. Given that quadrilateral $ALEX$ can be inscribed in a circle, we know that opposite angles of a cyclic quadrilateral sum to $180^\circ$. This is a property of cyclic quadrilaterals.
   
2. We are given $\angle LAX = 20^\circ$ and $\angle AXE = 100^\circ$.

3. Since $ALEX$ is a cyclic quadrilateral, the opposite angles $\angle LAX$ and $\angle LEX$ must sum to $180^\circ$. Therefore, we have:
   \[
   \angle LAX + \angle LEX = 180^\circ
   \]
   Substituting the given value:
   \[
   20^\circ + \angle LEX = 180^\circ \implies \angle LEX = 160^\circ
   \]

4. Next, we need to find $\angle EXD$. Since $\angle AXE = 100^\circ$, and the sum of angles around point $X$ is $360^\circ$, we have:
   \[
   \angle AXE + \angle EXD = 180^\circ
   \]
   Substituting the given value:
   \[
   100^\circ + \angle EXD = 180^\circ \implies \angle EXD = 80^\circ
   \]

5. Now, we need to find $\angle XED$. Since $\angle LEX = 160^\circ$, and the sum of angles around point $E$ is $360^\circ$, we have:
   \[
   \angle LEX + \angle XED = 180^\circ
   \]
   Substituting the value we found:
   \[
   160^\circ + \angle XED = 180^\circ \implies \angle XED = 20^\circ
   \]

6. Finally, we need to find $\angle EDX$. Using the fact that the sum of angles in a triangle is $180^\circ$, we consider triangle $EXD$:
   \[
   \angle EXD + \angle XED + \angle EDX = 180^\circ
   \]
   Substituting the values we found:
   \[
   80^\circ + 20^\circ + \angle EDX = 180^\circ \implies \angle EDX = 80^\circ
   \]

Therefore, the measure of $\angle EDX$ is $\boxed{80^\circ}$.