9. As shown in Figure 2, in quadrilateral $ABCD$, $AB=BC=CD$, $\angle ABC=78^{\circ}$, $\angle BCD=162^{\circ}$. Let the intersection point of line $AD$ and $BC$ be $E$. Then the size of $\angle AEB$ is
$9.21^{\circ}$.
As shown in Figure 5, draw $D O / / C B$, and
make $D O=C B$, connect
$O A$ and $O B$.
Since $B C=C D$,
thus, quadrilateral $B C D O$ is a rhombus. Therefore,
$$
\begin{array}{l}
A B=B C=B O . \\
\text { Since } \angle A B O=78^{\circ}-\left(180^{\circ}-162^{\circ}\right)=60^{\circ},
\end{array}
$$

Therefore, $\triangle A B O$ is an equilateral triangle.
Let $\angle A E B=\alpha$. Then $\angle 1=\angle 2=\alpha$.
By the sum of the interior angles of a quadrilateral being $360^{\circ}$, we get
$$
\begin{array}{l}
78^{\circ}+162^{\circ}+\left(60^{\circ}+\alpha\right)+\left(180^{\circ}-162^{\circ}+\alpha\right) \\
=360^{\circ} .
\end{array}
$$

Solving for $\alpha$ gives $\alpha=21^{\circ}$.