Problem 959

This problem has many proof methods, which have been widely explored by everyone. We will not repeat them here. By utilizing the conclusion of this competition problem and with the help of some well-known theorems, we obtain two new conclusions related to the excircles of a triangle as follows:

Proposition: Let $\triangle ABC$ be a non-equilateral triangle. The excircle opposite to $\angle A$ touches sides $AB$ and $AC$ at $A_3$ and $A_4$, respectively. The line $A_3A_4$ intersects line $BC$ at point $A_1$. Similarly, we can define $B_3, B_4, B_1$ and $C_3, C_4, C_1$. Furthermore, let the line $A_3A_4$ intersect $B_3B_4$ at point $C_2$, $B_3B_4$ intersect $C_3C_4$ at point $A_2$, and $C_3C_4$ intersect $A_3A_4$ at point $B_2$, as shown in Figure 2. Then
(1) The points $A_1, B_1, C_1$ are collinear;
(2) The circumcenter of $\triangle A_2B_2C_2$ is the orthocenter of $\triangle ABC$.

Proof: (1) From the conclusion of the original problem, we know that $A_2A \perp BC$, $B_2B \perp CA$, and $C_2C \perp AB$, i.e., $A_2A$, $B_2B$, and $C_2C$ are the altitudes of $\triangle ABC$. Therefore, $A_2A$, $B_2B$, and $C_2C$ concur at the orthocenter $H$, which means that the corresponding points of $\triangle ABC$ and $\triangle A_2B_2C_2$ are concurrent. Hence, by Desargues' Theorem, the points $A_1, B_1, C_1$ are collinear.
(2) We need to prove that $HA_2 = HB_2 = HC_2$. To prove $HB_2 = HC_2$, we need to prove $\angle HB_2C_2 = \angle HC_2B_2$. However, since $H$ is the orthocenter of $\triangle ABC$, we have $\angle HBA = \angle HCA$, i.e., $\angle B_2BA_3 = \angle C_2CA_4$. Furthermore, by the tangent segment theorem, we have $\angle AA_3A_4 = \angle AA_4A_3$, which implies $\angle HB_2C_2 = \angle HC_2B_2$. Therefore, the conclusion holds.