# Problem 9.3 (7 points)

$H$ is the point of intersection of the altitudes of an acute-angled triangle $ABC$. It is known that $HC = BA$. Find the angle $ACB$.

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# Solution:

Let $\mathrm{D}$ be the foot of the altitude dropped from vertex A to side BC. Angles НСВ and DAB are equal as acute angles with mutually perpendicular sides. Therefore, $\triangle \mathrm{CHD}=\Delta \mathrm{ABD}$ (by hypotenuse and acute angle). Hence, $=\mathrm{AD}$, i.e., triangle $\mathrm{ACD}$ is isosceles and right-angled, so $\angle \mathrm{ACB}=45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_118c47fc0d380e3dc110g-1.jpg?height=397&width=446&top_left_y=1977&top_left_x=1476)

| Criteria | Points |
| :--- | :---: |
| Complete solution | 7 |
| Proved $\triangle \mathrm{CHD}=\triangle \mathrm{ABD}$ | 5 |
| Proved $\angle \mathrm{HCB}=\angle \mathrm{DAB}$ | 3 |
| Correct answer without justification | 1 |

Answer: $\angle A C B=45^{\circ}$

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