5. As shown in Figure 2, in the plane, there are four line segments $A B, B C, C D, D A$ connected head to tail in sequence. Point $E$ is on the extension of $B A$, and the angle bisector of $\angle D A E$ intersects the angle bisector of $\angle B C D$ at point $F$. If $\angle B=26^{\circ}, \angle D=62^{\circ}$, then $\angle A F C=(\quad)$.
(A) $131^{\circ}$
(B) $132^{\circ}$
(C) $133^{\circ}$
(D) $134^{\circ}$
5. D.

According to the problem, let $\angle D A F=\angle E A F=\alpha$,
$$
\begin{array}{l}
\angle D C F=\angle B C F=\theta . \\
\text { Then } \angle B+180^{\circ}-2 \alpha=\angle D+2 \theta \\
\Rightarrow \alpha+\theta=90^{\circ}-\frac{1}{2}(\angle D-\angle B) .
\end{array}
$$

Therefore, $\angle A F C=\angle D+(\alpha+\theta)$
$$
=90^{\circ}+\frac{1}{2}(\angle D+\angle B)=134^{\circ} .
$$