12. In an acute triangle $\triangle ABC$, the midpoints of the three altitudes $AA_{1}, BB_{1}, CC_{1}$ are $A_{2}, B_{2}, C_{2}$, respectively. Find the sum of the angles $B_{2} A_{2} C_{2}, C_{2} B_{1} A_{2}, A_{2} C_{1} B_{2}$.
12. As shown in Figure 12-20, first find $\angle B_{2} A_{1} C_{2}$, connect $A_{1} B_{1}, A_{1} C_{1}$, extend $A_{1} B_{2}$ to $B^{\prime}{ }_{2}$ such that $B_{2} B^{\prime}{ }_{2}=A_{1} B_{2}$, extend $A_{1} C_{2}$ to $C^{\prime}{ }_{2}$ such that $C_{2} C^{\prime}{ }_{2}=A_{1} C_{2}$. It is easy to see that $\angle C^{\prime}{ }_{2} C_{1} A_{1}=\angle C_{1} A_{1} B=$ $\angle B^{\prime}{ }_{2} B_{1} A_{1}=\angle B_{1} A_{1} C=\angle A$ (using $C_{1} C^{\prime}{ }_{2} \parallel B C, B_{1} B_{2}{ }^{\prime} \parallel B C$, and the fact that the four points are concyclic), $B_{1} B^{\prime}{ }_{2}=A_{1} B, C_{1} C^{\prime}{ }_{2}=A_{1} C$. Also, $\triangle B A_{1} C_{1} \backsim \triangle B_{1} A_{1} C$, so $\frac{B A_{1}}{C_{1} A_{1}}=\frac{B_{1} A_{1}}{C A_{1}}$, hence $\frac{C_{1} C^{\prime}}{C_{1} A_{1}}=$ $\frac{B_{1} A_{1}}{B_{1} B^{\prime}}$, which means $\triangle C_{1} A_{1} C^{\prime}{ }_{2} \backsim \triangle B_{1} B^{\prime}{ }_{2} A$, thus $\angle B_{2} A_{1} C_{2}=\angle A$.
Similarly, $\angle C_{2} B_{1} A_{2}=\angle B, \angle A_{2} C_{1} B_{2}=\angle C$, therefore, the sum of these three angles is $\pi$.