1. Let $I$ denote the intersection of the three internal bisectors. Then $I A_{1}=A_{1} A^{0}$. One way proving this is to realize that the circumcircle of $A B C$ is the nine-point circle of $A^{0} B^{0} C^{0}$, hence it bisects $I A^{0}$, since $I$ is the orthocenter of $A^{0} B^{0} C^{0}$. Another way is through noting that $I A_{1}=A_{1} B$, which follows from $\angle A_{1} I B=\angle I B A_{1}=(\angle A+\angle B) / 2$, and $A_{1} B=A_{1} A^{0}$ which follows from $\angle A_{1} A^{0} B=\angle A_{1} B A^{0}=90^{\circ}-\angle I B A_{1}$. Hence, we obtain $S_{I A_{1} B}=S_{A^{0} A_{1} B}$. Repeating this argument for the six triangles that have a vertex at $I$ and adding them up gives us $S_{A^{0} B^{0} C^{0}}=2 S_{A C_{1} B A_{1} C B_{1}}$. To prove $S_{A C_{1} B A_{1} C B_{1}} \geq 2 S_{A B C}$, draw the three altitudes in triangle $A B C$ intersecting in $H$. Let $X, Y$, and $Z$ be the symmetric points of $H$ with respect to the sides $B C, C A$, and $A B$, respectively. Then $X, Y, Z$ are points on the circumcircle of $\triangle A B C$ (because $\left.\angle B X C=\angle B H C=180^{\circ}-\angle A\right)$. Since $A_{1}$ is the midpoint of the arc $B C$, we have $S_{B A_{1} C} \geq S_{B X C}$. Hence
$$S_{A C_{1} B A_{1} C B_{1}} \geq S_{A Z B X C Y}=2\left(S_{B H C}+S_{C H A}+S_{A H B}\right)=2 S_{A B C}$$