5. As shown in Figure 3, quadrilateral $ABCD$ is a rhombus, $\triangle AEF$ is an equilateral triangle, points $E$ and $F$ are on sides $BC$ and $CD$ respectively, and $AB = AE$. Then $\angle B$ equals ( ).
(A) $60^{\circ}$
(B) $80^{\circ}$
(C) $100^{\circ}$
(D) $120^{\circ}$
5.B.

Since quadrilateral $A B C D$ is a rhombus, $A B=A D, \angle B=\angle D$.
Also, since $\triangle A E F$ is an equilateral triangle, then $A E=A F$. Therefore, $\triangle A B E \cong \triangle A D F$. Hence $B E=D F$.
Since $B C=C D$, then $C E=C F$. Therefore, $\angle C E F=\frac{1}{2}\left(180^{\circ}-\angle C\right)$.
From $A B=A E$, we know $\angle A E B=\angle B$.
And $\angle A E B+\angle A E F+\angle C E F=180^{\circ}$,
and $\angle C+\angle B=180^{\circ}, \angle A E F=60^{\circ}$, so $\angle B+\frac{1}{2} \angle B=120^{\circ}$. Therefore, $\angle B=80^{\circ}$.