7.1. Mihai drew several right-angled triangles on the board, such that any two triangles do not have common points. In each of these triangles, Mihai drew the following important lines (segments): all angle bisectors, all medians, and the altitude to the hypotenuse. Counting, the total number of important lines drawn in all the triangles is 44. Determine how many right-angled isosceles triangles Mihai drew on the board.
## Solution:

Right-angled triangles can be scalene or isosceles. In a scalene right-angled triangle, Mihai constructs exactly 7 distinct important segments: 3 angle bisectors, 3 medians, and one altitude. In a right-angled isosceles triangle, Mihai constructs exactly 5 distinct important segments, because the median, altitude, and angle bisector drawn from the right angle vertex coincide. We note that no important segment in one triangle coincides with any important segment in another triangle. Let $x$ be the number of any right-angled triangles, and $y$ be the number of right-angled isosceles triangles drawn by Mihai. Thus, we get $7 x + 5 y = 44$, where $x, y \in \mathbb{N}$. Since $7 x \leq 44$ and $x \in \mathbb{N}$, it follows that $x \in \{0,1,2,3,4,5,6\}$.

For $x=0$, we get $y=\frac{44}{5} \notin \mathbb{N}$. For $x=1$, we get $y=\frac{37}{5} \notin \mathbb{N}$. For $x=2$, we get $y=6$. For $x=3$, we get $y=\frac{23}{5} \notin \mathbb{N}$. For $x=4$, we get $y=\frac{16}{5} \notin \mathbb{N}$. For $x=5$, we get $y=\frac{9}{5} \notin \mathbb{N}$. For $x=6$, we get $y=\frac{2}{5} \notin \mathbb{N}$. Since the only solution to the equation $7 x + 5 y = 44$, where $x, y \in \mathbb{N}$, is $x=2$ and $y=6$, it follows that Mihai drew 6 right-angled isosceles triangles.

Answer: Mihai drew 6 right-angled isosceles triangles.