5. Given a convex hexagon $A B C D E F$ with six equal side lengths, the interior angles $\angle A$, $\angle B$, and $\angle C$ are $134^{\circ}$, $106^{\circ}$, and $134^{\circ}$, respectively. Then the measure of the interior angle $\angle E$ is $\qquad$.
5. $134^{\circ}$.

Let's assume the side length is $1$, and the midpoints of $AC$ and $DF$ are $M$ and $N$, respectively, and the projection of point $A$ on $DF$ is $K$. Then
$$
\begin{array}{l}
\angle B A M=37^{\circ}, \angle M A F=97^{\circ}, \\
\angle A F K=83^{\circ} .
\end{array}
$$

Thus, $F K=\cos 83^{\circ}, K N=A M=\cos 37^{\circ}$.
Let $\angle E F N=x$. Then $F N=\cos x$.
Using $F N=F K+K N$, we get
$$
\begin{array}{l}
\cos x=\cos 83^{\circ}+\cos 37^{\circ} \\
=2 \cos 60^{\circ} \cdot \cos 23^{\circ}=\cos 23^{\circ} .
\end{array}
$$

Therefore, $x=23^{\circ}$, which means the base angle of isosceles $\triangle D E F$ is $23^{\circ}$.
Thus, its vertex angle $\angle E=134^{\circ}$.