6. As shown in Figure 3, let $I$ be the incenter of $\triangle A B C$. If $\angle A B C=35^{\circ}$, and $B C$ $=A I+A C$, then $\angle B A C$ $=(\quad)$.
(A) $55^{\circ}$
(B) $60^{\circ}$
(C) $65^{\circ}$
(D) $70^{\circ}$
6. D.

As shown in Figure 6, on $BC$, intercept $CD = CA$, and connect $ID$. Then
$\triangle ACI \cong \triangle DCI$.
Thus, $AI = DI$,
$\angle CDI = \angle CAI$.
Since $BC = AI + AC$
$$
= CD + BD \text{, }
$$

Therefore, $BD = AI = DI$.
Hence, $\angle BID = \angle IBD = \angle ABI$.
Thus, $\angle BAC = 2 \angle IDC = 2 \angle ABC = 70^{\circ}$.