5. As shown in Figure 2, points $E$ and $F$ are on the sides $AB$ and $BC$ of rhombus $ABCD$, respectively, and $\frac{AE}{BE} = \frac{BF}{CF}$. If $\triangle DEF$ is an equilateral triangle, then the degree measure of $\angle BAD$ is ( ).
(A) $40^{\circ}$
(B) $60^{\circ}$
(C) $80^{\circ}$
(D) $100^{\circ}$
5.B.

As shown in Figure 7, from the problem, we know that $B E = C F$. Take a point $K$ on side $A B$ such that $A K = B E$. It is easy to see that
$\triangle A K D \cong \triangle C F D$.
Thus, $D K = D F = D E$.
Therefore, $\triangle D K E$ is an isosceles triangle, and we have
$$
\angle D K E = \angle D E K \text{. }
$$

Then $\triangle D K A \cong \triangle D E B$.
Hence, $A D = B D$. Therefore, $\triangle A B D$ is an equilateral triangle.