2. In triangle $ABC$, the altitudes $AA_{0}$, $BB_{0}$, and $CC_{0}$ are dropped. Let $H$ be the orthocenter. If $\overline{A H}: \overline{H A_{0}}=1: 1$ and $\overline{B H}: \overline{H B_{0}}=2: 1$, then what is the ratio $\overline{C H}: \overline{H C_{0}}$?
Solution. Note that

![](https://cdn.mathpix.com/cropped/2024_06_05_0537e7aa6f770cbd33c5g-1.jpg?height=128&width=380&top_left_y=1897&top_left_x=341)

i.e., \( P_{\triangle B C H} = \frac{1}{2} P_{\triangle A B C} \). Similarly,

\[
\frac{P_{\triangle A B C}}{P_{\triangle A C H}} = \frac{\overline{B B_{0}}}{\overline{H B_{0}}} = 3
\]

![](https://cdn.mathpix.com/cropped/2024_06_05_0537e7aa6f770cbd33c5g-1.jpg?height=439&width=599&top_left_y=1804&top_left_x=883)
i.e., \( P_{\triangle A C H} = \frac{1}{3} P_{\triangle A B C} \).

Note that

\[
P_{\triangle A B C} = P_{\triangle B C H} + P_{\triangle A C H} + P_{\triangle A B H}
\]

from which we get \( P_{\triangle A B H} = \frac{1}{6} P_{\triangle A B C} \). From this, we have

\[
6 = \frac{P_{\triangle A B C}}{P_{\triangle A B H}} = \frac{\overline{C C_{0}}}{\overline{H C_{0}}}
\]

i.e., \( \overline{C H} : \overline{H C_{0}} = 5 : 1 \).