2. In an acute triangle $\triangle ABC$, two altitudes $BY'$ and $CC'$ intersect at $H$. If $AH + AB' + AC' = BC + BB' + CC'$, then $\angle BAC = (\quad)$.
(A) $75^{\circ}$
(B) $60^{\circ}$.
(C) $45^{\circ}$
(D) $30^{\circ}$
2. (C).

It is easy to know $\triangle A H B^{\prime} \backsim \triangle B C B^{\prime}, \triangle A B B^{\prime} \backsim \triangle A C C^{\prime}$
$$
\begin{array}{l}
\Rightarrow \frac{A H}{B C}=\frac{A B^{\prime}}{B B^{\prime}}=\frac{A C^{\prime}}{C C^{\prime}}=\frac{A H+A B^{\prime}+A C^{\prime}}{B C+B B^{\prime}+C C^{\prime}}=1 \\
\Rightarrow A B^{\prime}=B B^{\prime}\left(\text { or } A C^{\prime}=\left(C^{\prime}\right) \Rightarrow \angle B A C=45^{\circ} .\right.
\end{array}
$$