Let $H$ be the orthocenter of a triangle $ABC$, and $H_{A}, H_{B}, H_{C}$ the feet of the altitudes from $A, B, C$. For simplicity, we assume that $H$ is inside the triangle $ABC$, which means that all angles of the triangle are acute (such a triangle is called acute-angled). Determine the angles of the triangles $A H_{B} H_{C}, H_{A} B H_{C}, H_{A} H_{B} C$ and $H_{A} H_{B} H_{C}$, in terms of the angles of the triangle $ABC$, which we will denote as $\widehat{A}, \widehat{B}$ and $\widehat{C}$.

Note: In triangle geometry, many points are used, and to avoid using the same name for too many different points, it can happen that one runs out of notations. The notations $H_{A}, H_{B}$ and $H_{C}$ are not common, but they can be useful in many cases.

![](https://cdn.mathpix.com/cropped/2024_05_10_831e903be1b8a1180f82g-29.jpg?height=610&width=748&top_left_y=535&top_left_x=407)
Given the right angles $\widehat{B H_{B} C}$ and $\widehat{B H_{C} C}$, $H_{B}$ and $H_{C}$ are on the circle with diameter $[B C]$, hence the inscribed angles $\widehat{B H_{C} H_{B}}$ and $\widehat{B C H_{B}}$ are supplementary, since $C$ and $H_{C}$ are on either side of $\left(B H_{B}\right)$, thus $\widehat{A H_{C} H_{B}}=\widehat{C}$. Similarly, $\widehat{A H_{B} H_{C}}=\widehat{B}$, then $\widehat{B H_{C} H_{A}}=\widehat{C}, \widehat{B H_{A} H_{C}}=\widehat{A}=\widehat{C H_{A} H_{B}}$ and $\widehat{C H_{B} H_{A}}=\widehat{B}$. Therefore, on the one hand, $\widehat{H_{A} H_{B} H_{C}}=180^{\circ}-2 \widehat{B}, \widehat{H_{B} H_{C} H_{A}}=180^{\circ}-2 \widehat{C}, \widehat{H_{C} H_{A} H_{B}}=180^{\circ}-$ $2 \widehat{A}$, on the other hand, the altitudes of triangle $A B C$ are the angle bisectors of triangle $H_{A} H_{B} H_{C}$, and the orthocenter of $A B C$ is the center of the inscribed circle in $H_{A} H_{B} H_{C}$.