<s>[INST] An interesting point of a triangle. (Help needed to prove a statement.) 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.
 [/INST] Any two ideal triangles are congruent, or in other words, any non-degenerate triangle inscribed into the unit circle can be mapped to any other by a projective transformation which fixes the unit circle. So without loss of generality you can restrict your considerations to one special case.
I like coordinates, so I'd start with $\triangle A_1A_2A_3$, the matrix $M$ of its circumcircle, and point $P$ chosen as follows (in homogeneous coordinates):
$$P=\begin{pmatrix}x\\y\\z\end{pmatrix} \qquad
M=\begin{pmatrix}1&&\\&1&\\&&-1\end{pmatrix}$$
\begin{align*}
A_1&=\begin{pmatrix}1\\0\\1\end{pmatrix} &
A_2&=\begin{pmatrix}0\\1\\1\end{pmatrix} &
A_3&=\begin{pmatrix}0\\-1\\1\end{pmatrix} \\
a_1=A_2\vee A_3&=\begin{pmatrix}1\\0\\0\end{pmatrix} &
a_2=A_3\vee A_1&=\begin{pmatrix}1\\-1\\-1\end{pmatrix} &
a_3=A_1\vee A_2&=\begin{pmatrix}1\\1\\-1\end{pmatrix} \\
B_1=Ma_1&=\begin{pmatrix}1\\0\\0\end{pmatrix} &
B_2=Ma_2&=\begin{pmatrix}1\\-1\\1\end{pmatrix} &
B_3=Ma_3&=\begin{pmatrix}1\\1\\1\end{pmatrix} \\
b_1=P\vee B_1&=\begin{pmatrix}0\\z\\-y\end{pmatrix} &
b_2=P\vee B_2&=\begin{pmatrix}y+z\\z-x\\-x-y\end{pmatrix} &
b_3=P\vee B_3&=\begin{pmatrix}y-z\\z-x\\x-y\end{pmatrix} \\
C_1=a_1\wedge b_1&=\begin{pmatrix}0\\y\\z\end{pmatrix} &
C_2=a_2\wedge b_2&=\begin{pmatrix}z+y\\x-z\\2z-x+y\end{pmatrix} &
C_3=a_3\wedge b_3&=\begin{pmatrix}z-y\\z-x\\2z-x-y\end{pmatrix} \\
c_1=A_1\vee C_1&=\begin{pmatrix}y\\z\\-y\end{pmatrix} &
c_2=A_2\vee C_2&=\begin{pmatrix}3z-2x+y\\y+z\\-y-z\end{pmatrix} &
c_3=A_3\vee C_3&=\begin{pmatrix}3z-2x-y\\y-z\\y-z\end{pmatrix}
\end{align*}
$A_i$ are the corners of the triangle. $a_i$ its edges, your solid colored lines. $B_i$ their poles, your $\color{red}P,\color{green}P,\color{blue}P$. $b_i$ the connection of these to $P$, i.e. the perpendiculars, which you drew using dashed lines. $C_i$ are the points where these intersect the edges. Finally, $c_i$ are the lines connecting these, which you drew in white.
These three lines $c_i$ are concurrent if their determinant is zero.
$$\begin{vmatrix}
y&3z-2x+y&3z-2x-y\\
z&y+z&y-z\\
-y&-y-z&y-z
\end{vmatrix}=0$$
The point of concurrency is
$$Q=c_1\wedge c_2=
\begin{pmatrix}y^2- z^2 \\ 2xy - 2yz \\ y^2 + 2xz - 3z^2\end{pmatrix}$$
There are some degenerate situations, where that point becomes undefined, i.e. the null vector. Computing all solutions to this set of three quadratic homogeneous equations, you find the degenerate situations to be
$$P\in\left\{
\begin{pmatrix}1\\0\\0\end{pmatrix},
\begin{pmatrix}1\\-1\\1\end{pmatrix},
\begin{pmatrix}1\\1\\1\end{pmatrix}
\right\}=\{B_1,B_2,B_3\}$$
Those are the cases you explicitely excluded, so in all other situations, everything is fine.
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