3. In $\triangle A B C$, $\angle B=100^{\circ}$, the angle bisector of $\angle C$ intersects side $A B$ at $E$, take point $D$ on side $A C$ such that $\angle C B D=20^{\circ}$, and connect $D, E$. Then the degree measure of $\angle C E D$ is
3. Let the reverse extension line of $CB$ be $BF$, then $\angle ABF=80^{\circ}$, $\angle ABD=80^{\circ}$. Therefore, $BA$ is the bisector of $\angle DBF$. Since $E$ is on $BA$, the distances from $E$ to $BF$ and $BD$ are equal.

Since $E$ is on the bisector $CE$ of $\angle ACB$, the distances from $E$ to $CF$ and $CA$ are equal.

From the above, the distances from $E$ to the two sides of $\angle ADB$ are equal, so $E$ lies on the bisector of $\angle ADB$. Thus, $\angle ADE=\frac{1}{2} \angle ADB$. Therefore, $\angle CED=\angle ADE-\angle ACE=10^{\circ}$.