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 $\boxed{80^\circ}$.