Let $G$ is circumcenter of $ABD$. Then $G$ lies on perpendicular bisector of $AB$ as $X$ and $Y$ so $G$ lies on $XY$. Also $G$ lies on perpendicular bisector of $AD$ and $T,Z$ too, so $G$ is point of intersection of $XY$ and $TZ$. $X$ lies on perpendicular bisector to $EB$ and as $BC=CD=DE$ so $X$ lies on perpendicular bisector to $CD$, $G$ lies on perpendicular bisector to $BD$ and $Y$ lies on perpendicular bisector to $BC$, and $BC=CD$ so $G$ is midpoint of $XY$. So $GZ$ is median of $XYZ$ and $T$ lies on median. Same way we can prove, that $T$ lies on other median of $XYZ$, so $T$ is centroid of $XYZ$