G26 (9-4, Italy) Given two acute triangles $\triangle A_{0} B_{0} C_{0}$ and $\triangle A_{1} B_{1} C_{1}$, construct a triangle $\triangle A B C$ such that $\triangle A B C \backsim \triangle A_{1} B_{1} C_{1}$ (with $A, B, C$ corresponding to $A_{1}$, $B_{1}$, $C_{1}$ respectively), and $\triangle A B C$ is circumscribed around $\triangle A_{0} B_{0} C_{0}$ (with $A_{0}, B_{0}, C_{0}$ lying on $B C, A C$, $A B$ respectively), and then construct the triangle that satisfies the above conditions and has the maximum area.
Solve: Construct arcs with $A_{0} C_{0}$ and $A_{0} B_{0}$ as chords of $\triangle A_{0} B_{0} C_{0}$, and $\angle B_{1}$ and $\angle C_{1}$ as inscribed angles, as shown in Figure 426.

Take points $B$ and $C$ on the two arcs respectively, such that $B, A_{0}, C$ are collinear. Extend $B C_{0}$ and $C B_{0}$ to intersect at $A$. It is easy to see that $\angle A = \angle A_{1}$, thus $\triangle A B C \backsim \triangle A_{1} B_{1} C_{1}$. It is evident that $B$ can be any point on $\overparen{A_{0} C_{0}}$, and $C$ depends on $B$.

As shown in the figure, let the centers of the two arcs be $O_{1}$ and $O_{2}$ respectively. Draw $O_{1} P_{1}$ and $O_{2} P_{2}$ perpendicular to $B C$, with $P_{1}$ and $P_{2}$ as the feet of the perpendiculars. Thus, $P_{1}$ and $P_{2}$ are the midpoints of $B A_{0}$ and $A_{0} C$ respectively, and therefore
$$
B C = 2 P_{1} P_{2} \leqslant 2 O_{1} O_{2}.
$$

Equality holds if and only if $B C \parallel O_{1} O_{2}$. From this, it is not difficult to determine the exact position of the triangle that satisfies the condition and is the largest.