Consider a triangle whose sides are segments of $\color{red}{\text{line}}$, $\color{blue}{\text{line}}$, $\color{green}{\text{line}}$ falling in the circum-circle $c$. Let $\color{red}{\text{P}}$,$\color{green}{\text{P}}$, $\color{blue}{\text{P}}$ be the poles (with respect to $c$) of the corresponding sides of the triangle.
Now, take a point $P$ different from the poles. Connect the poles with $P$. The connecting lines will intersect the corresponding edges or the elongations of theses edges mentioned above (perhaps in the $\infty$). (Corresponding means: $\color{red}{\text{ red broken line}}$ with $\color{red}{\text{ red edge line }}$, etc.
Then connect the vertices of the triangle with the the opposite intersection points mentioned above as shown in the figure below (white lines). The white lines will meet in one point. (Perhaps in the infinity; then the white lines are parallel.)
I call this point the $P$-pole point of the triangle with respect to its circum-circle and point $P$. I cannot prove that the pole point always exists. (It exists even if the white lines are parallel.) Any help, please? Any known results?
The same statement can be told easier in the language of hyperbolic geometry: Take an ideal triangle and a point $P$ not on the sides. Drop perpendiculars from $P$ to the sides of the triangle. Consider the intersection points. Then connect these intersection points with the opposite vertices with suitable parallels. These parallels will meet in one point, the "pole point of the ideal triangle-with respect to $P$. (See the figure below.)
To be honest I don't have a clue as to how to prove the statements given above. I found the "pole point" in the clear blue.