5. In $\triangle A B C$, it is known that $A B=A C$, point $D$ is on side $A C$, $\angle A D B=60^{\circ}$, point $E$ is on side $B D$, $\angle E C B=30^{\circ}$. Find the degree measure of $\angle A E C$.
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Let point $F$ be the reflection of point $E$ over $BC$.
\[
\begin{array}{l}
\text { Then } \angle B F C=\angle B E C \\
=180^{\circ}-\frac{1}{2} \angle B A C .
\end{array}
\]

By criterion 3, $A$ is the circumcenter of $\triangle B C F$.
It is also easy to prove that $\triangle C E F$ is an isosceles triangle.
Thus, $\triangle A E F \cong \triangle A E C$.
Answer: $150^{\circ}$.
```