\documentclass[11pt]{scrartcl}
\usepackage{evan}

\begin{document}
\title{Twitch 031.1}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 31}
\maketitle

\section*{Problem}
$\triangle BB'C$ is inscribed in circle $\omega$ with center $A$
($\angle BCB'=90^{\circ}, \angle CB'B\geq \angle CBB'$).
Line $\alpha$ passes through $A$ and is perpendicular to $CA$.
Line $\alpha$ intersects circle $\omega$ at $D$ and $E$,
where $D$ lies on minor arc $BC$.
The intersection of ray $B'C$ through $C$ and the angle bisector of $\angle B'ED$ is $F$.
Line $FD$ intersects circle $\omega$ at $H\neq D$.
Show that line $HA$ is the perpendicular bisector of $\ol{BD}$.

\section*{Video}
\href{https://www.youtube.com/watch?v=6ou11s8WP0A&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/6ou11s8WP0A}}



\newpage

\section*{Solution}
The point $F$ is the $E$-excenter of $\triangle EB'D$.

\begin{center}
\begin{asy}
/* start ggb to asy preamble */
import graph; size(3.43465cm);
pen zzttqq = rgb(0.6,0.2,0.);
pair A = (0.,0.), B = (-0.67881,-0.73430), C = (0.,1.), D = (-1.,0.), F = (-1.31693,1.51545), H = (-0.91619,-0.40073);
/* end preamble */
draw((0.67881,0.73430)--(1.,0.)--D--cycle, linewidth(0.6) + zzttqq);
draw(circle(A, 1.), linewidth(0.6));
draw(B--(0.67881,0.73430), linewidth(0.6));
draw((1.,0.)--F, linewidth(0.6));
draw((0.67881,0.73430)--F, linewidth(0.6));
draw(F--H, linewidth(0.6));
draw((0.67881,0.73430)--(1.,0.), linewidth(0.6) + zzttqq);
draw((1.,0.)--D, linewidth(0.6) + zzttqq);
draw(D--(0.67881,0.73430), linewidth(0.6) + zzttqq);
dot("$A$", A, dir((2.391, -12.304)));
dot("$B$", B, dir((-6.631, -9.209)));
dot("$B'$", (0.67881,0.73430), dir((1.014, 2.115)));
dot("$C$", C, dir((1.022, 2.639)));
dot("$D$", D, dir((-10.637, -2.726)));
dot("$E$", (1.,0.), dir((3.925, -1.357)));
dot("$F$", F, dir((1.078, 2.272)));
dot("$H$", H, dir((-10.534, -8.357)));
dot("$N$", (-0.40073,0.91619), dir((1.140, 2.263)));
\end{asy}
\end{center}

The end.

\end{document}
