Show that U,V and H are colinear We are given a regular icosagon as below:

I wanna prove that the red line exists.
I know that $U$ is the incenter of $\triangle TLB$ ($T,U,G$ are collinear)
I know that $V$ is the incenter of $\triangle GMC$ ($M,V,E$ are collinear)
I tried to use Pascal theorem but it probably has to be used more than once and I have no clue which one would be useful... Maybe letting line $UV$ hit the circle in two random points may help but I could not develop it. There are 6 pairs of points to choose from in order to get the Pascal hexagons.
 A: We’ll use your observation that $V = CJ ∩ GR ∩ ME$ is the incenter of $\triangle CGM$. Let $W = BP ∩ CJ$, $Z = EM ∩ AL$. Translation by $\overrightarrow{BM} = \overrightarrow{CL}$ sends lines $BP, CJ$ to lines $EM, AL$, so it also sends $W$ to $Z$. Now $\triangle UWZ$ and $\triangle HJE$ have parallel corresponding sides, which means they are homothetic about $V$. Hence $U, V, H$ are collinear.

A: Here is a solution using complex numbers, i was waiting for the bounty to be given, myself having no simple synthetic solution.
I am still posting it, since there was an interesting experience of computing in the cyclotomic field $\Bbb Q(\zeta_{20})$.
Let $z=\zeta_{20}=\cos\frac{2\pi}{20}+i\cos\frac{2\pi}{20}$ be this primitive root of order $20$, a root of the cyclotomic polynomial $\Phi_{20}=Z^8-Z^6+Z^4-Z^2+1$.
We identify the vertices $A,B,\dots,T$ of the given $20$-gon with their affixes $a, b, \dots,t$ which are of the shape $z^k\in \Bbb C$, $k\in\Bbb Z/20$, so that $a=1$, $b=z$, $c=z^2$, and so on.
Then $V$ is the orthocenter of $\Delta JRE$, and the centroid is between zero and $V$ at one third from zero, so
$v=3\cdot\frac 13(j+r+e)=z^9+z^{17}+z^4 = z^4+z^9-z^7$. (Or use the parallelogram $VEHJ$.)
Then $U$ is the incenter of the right, isosceles $\Delta TLB$, it can be obtained as a $+90^\circ$-rotation of $B$ around $A$, so $u = 1 + z^5(z-1)$.
Then $U,V,H$ are collinear, iff the fraction $(u-h)/(v-h)$ is real, so iff the product $(u-h)(\bar v  - \bar h)$ is invariated by complex conjugation. Now we compute:
$$
\begin{aligned}
(u-h)
&=(z^6-z^5+1-z^7)=(z^6+1)-z^5(z^2+1)\\
&=(z^2+1)(z^4-z^2+1-z^5)=-(z^2+1)(z-1)(z^4+z+1)
\\[2mm]
(\bar v-\bar h) 
&=\overline{z^9 - 2z^7 + z^4}
=z^5(z-1)(z^5+1)(z^3-z-1)\ ,
\\[2mm]
(u-h)(\bar v-\bar h) 
&=-z^5\cdot (z-1)^2\cdot (z^2+1)\cdot (z^5+1)
\cdot \underbrace{(z^3-z-1)(z^4+z+1)}_{=-\frac 1z(z^3+z^2+z+1)}\\
&= z^4\cdot (z-1)\cdot (z^2+1)\cdot (z^5+1)\cdot (z^4-1) 
\ ,
\end{aligned}
$$
and the last expression is real, i.e. invariated by complex conjugation, because for $z^k\ne -1$
$$
\frac{1+z^k}{\overline{1+z^k}}=z^k\text{ , so }
\frac{\ z  -1\ }{\overline{z-1}}=-z\ ,\
\frac{\ z^2+1\ }{\overline{z^2+1}}=z^2\ ,\
\frac{\ z^5+1\ }{\overline{z^5+1}}=z^5\ ,\
\frac{\ z^4-1\ }{\overline{z^4-1}}=-z^4\ .
$$
$\square$
A: W.l.g. let the circumcircle of the icosagon have centre the origin $O$ and radius $1$. Now calculate several cartesian coordinates (which is made easier since $ME$ and $PB$ are at $45$ degrees to the axes).
$H$ is $(\cos 27, \sin 27)$.
$V$ is the intersection of $x=\sin27$ with the line through $M(-\sin27,\cos27)$ and $E(\cos27,-\sin27)$. So we have $V(\sin27,\cos27-2\sin27)$.
$U$ is the intersection of $x=-\sin9$ with the line through $P(-\cos9,\sin9)$ and $B(\sin9,-\cos9)$. So we have $U(-\sin9,2\sin9-\cos9)$.
$H,U,V$ are collinear if
$$(3\sin27-\cos27)(\sin9+\cos27)=(\sin27+\cos9-2\sin9)(\cos27-\sin27)$$ Equivalently
$$2\sin27\cos27+\sin9\cos27+\cos9\sin27=\cos ^227-\sin ^227+\cos9\cos27-\sin9\sin27$$ or
$$\sin 54+\sin36=\cos54+\cos36.$$
This is clearly true since $\sin54=\cos36$ and $\sin36=\cos54$.
