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2025 Macedonian Balkan MO TST, 2

Let $\triangle ABC$ be an acute-angled triangle and $A_1, B_1$, and $C_1$ be the feet of the altitudes from $A, B$, and $C$, respectively. On the rays $AA_1, BB_1$, and $CC_1$, we have points $A_2, B_2$, and $C_2$ respectively, lying outside of $\triangle ABC$, such that \[\frac{A_1A_2}{AA_1} = \frac{B_1B_2}{BB_1} = \frac{C_1C_2}{CC_1}.\] If the intersections of $B_1C_2$ and $B_2C_1$, $C_1A_2$ and $C_2A_1$, and $A_1B_2$ and $A_2B_1$ are $A', B'$, and $C'$ respectively, prove that $AA', BB'$, and $CC'$ have a common point.