TITLE: An interesting point of a triangle. (Help needed to prove a statement.)
QUESTION [5 upvotes]: 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.

REPLY [0 votes]: Here I am going to prove the first formulation of the statement in completely non-analytic way. In this proof, by "sides" of a triangle I will mean lines containing these sides.
I will first prove a lemma: let $A_1A_2A_3$ is any triangle and let $B_1,B_2,B_3$ be points on sides of this triangle opposite to $A_1,A_2,A_3$ respectively, such that $A_1B_1,A_2B_2,A_3B_3$ are concurrent at point $P$. Also let $C_1,C_2,C_3$ lie on sides of triangle $B_1B_2B_3$ such that $A_1C_1,A_2C_2,A_3C_3$ are concurrent at point $Q$. Then lines $B_1C_1,B_2C_2,B_3C_3$ are concurrent.
http://puu.sh/hLPba/0fdcff667a.png
(NOTE: we don't know yet if red point is a point of concurrence)
Proof of lemma: Let $X_i$ be the intersection of line $A_iB_i$ with side of triangle $B_1B_2B_3$, and let $Y_i$ be the intersection of line $A_iC_i$ with side of triangle $A_1A_2A_3$. Also let's denote by $[XYZ]$, for collinear points $X,Y,Z$, value $\frac{XY}{YZ}$ if $Y$ is between $X,Z$ and $-\frac{XY}{YZ}$ otherwise. We can formulate Ceva's theorem as "$AK,BL,CM$ are concurrent cevians of $\triangle ABC$ iff $[BKC]\cdot[CLA]\cdot[AMB]=1$.
By projective invariance of cross-ratio we have $\frac{[A_1B_3A_2]}{[A_1Y_3A_2]}=\frac{[B_2X_3B_1]}{[B_2C_3B_1]},\frac{[A_2B_1A_3]}{[A_2Y_1A_3]}=\frac{[B_3X_1B_2]}{[B_3C_1B_2]},\frac{[A_3B_2A_1]}{[A_3Y_2A_1]}=\frac{[B_1X_2B_3]}{[B_1C_2B_3]}$ so by multiplying we get $\frac{[A_1B_3A_2]}{[A_1Y_3A_2]}\frac{[A_2B_1A_3]}{[A_2Y_1A_3]}\frac{[A_3B_2A_1]}{[A_3Y_2A_1]}=\frac{[B_2X_3B_1]}{[B_2C_3B_1]}\frac{[B_3X_1B_2]}{[B_3C_1B_2]}\frac{[B_1X_2B_3]}{[B_1C_2B_3]}$. Now by Ceva's theorem, numerators of both sides and denominator of LHS are all equal to $1$, so denominator of RHS is equal to $1$, so again by Ceva $A_1Y_1,A_2Y_2,A_3Y_3$ are concurrent and lemma follows.
Now we want to find this kind of configuration in your problem. If we take colorful points $P$ from your problem as $A_1,A_2,A_3$ in the lemma and vertices of your triangle as $B_1,B_2,B_3$ then we have that $A_1B_1,A_2B_2,A_3B_3$ are concurrent by using certain degenerated form of Brianchon's theorem (note that the circle in question is tangent to the sides of triangle formed by colorful points $P$, because pole of a line intersecting a circle is intersection of tangents through intersection points). Now black point $P$ from your problem will work like point $Q$ from lemma, if we take $C_1,C_2,C_3$ to be colorful points on triangle's sides. By lemma, lines $B_1C_1,B_2C_2,B_3C_3$ will be concurrent - but these are precisely the white lines you ask about. So the pole indeed always exists (except for the cases when black $P$ is at one of triangle's sides).
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