Solution. a) From the conditions of the problem, we have:

$$
P_{A H C}=\frac{1}{2} \overline{A C} \cdot \frac{1}{3} \overline{B E}=\frac{1}{3} P_{A B C} \quad \text { and } \quad P_{B H C}=\frac{1}{2} \overline{B C} \cdot \frac{1}{2} \overline{A D}=\frac{1}{2} P_{A B C}
$$

Therefore,

$$
\begin{aligned}
P_{A H B} & =P_{A B C}-P_{A H C}-P_{B H C} \\
& =P_{A B C}-\frac{1}{2} P_{A B C}-\frac{1}{3} P_{A B C}=\frac{1}{6} P_{A B C}
\end{aligned}
$$

However,

$$
P_{A H B}=\frac{1}{2} \overline{A B} \cdot \overline{H F} \quad \text { and } P_{A B C}=\frac{1}{2} \overline{A B} \cdot \overline{C F},
$$

and if we substitute in the last equality, we get $\overline{H F}=\frac{1}{6} \overline{C F}$, from which it follows that $\overline{C H}: \overline{H F}=5: 1$.

b) Let $G$ be the midpoint of $\overline{B H}$. Triangle $B D H$

![](https://cdn.mathpix.com/cropped/2024_06_05_6b1323df06a0e4d4e36cg-3.jpg?height=447&width=391&top_left_y=228&top_left_x=1088)
is a right triangle, with a right angle at vertex $D$, so

$$
\overline{B G}=\overline{G H}=\overline{G D}
$$

From $\overline{A H}=\overline{H D}, \overline{E H}=\overline{H G}$ and $\measuredangle A H E=\measuredangle D H G$, it follows that $\triangle A E H \cong \triangle D G H$. Therefore, $\overline{A E}=\overline{D G}=\overline{H G}=\overline{E H}$, i.e., $\triangle A E H$ is an isosceles right triangle, so $\measuredangle E A H=45^{\circ}$. Finally, from triangle $C A D$ we get

$$
\measuredangle A C B=90^{\circ}-\measuredangle C A D=90^{\circ}-\measuredangle E A H=45^{\circ}
$$

## 2nd Year
4. Let $H$ be the orthocenter of the acute-angled triangle $ABC$, and let $D, E$, and $F$ be the feet of the altitudes dropped from vertices $A, B$, and $C$ respectively, and let $\overline{A H}=\overline{H D}$ and $\overline{B H}=2 \overline{H E}$.

a) Find the ratio $\overline{C H}: \overline{H F}$,

b) Find the angle $A C B$.