5. As shown in Figure 1, in $\triangle A B C$, $C B>C A, \angle B A C$ $=80^{\circ}, D$ is a point on $A B$ such that $C B-C A=B D$, $I$ is the intersection of the three angle bisectors of $\triangle A B C$. Then $\angle I D A=$ ( ).
(A) $30^{\circ}$
(B) $40^{\circ}$
(C) $50^{\circ}$
(D) $60^{\circ}$
5. B.

As shown in Figure 7, take point $A_{1}$ on $CB$ such that $CA_{1} = CA$. Connect $IA_{1}$, $IA$, $A_{1}D$, $CI$, and $BI$. Then $BD = BA_{1}$.
Figure 7

It is easy to prove that $\triangle ACI \cong \triangle A_{1}CI$,
$\triangle A_{1}BI \cong \triangle DBI$.
Therefore, $AI = A_{1}I = DI$.
Thus, $\angle IDA = \angle IAD = \frac{1}{2} \angle DAC = 40^{\circ}$.