Let $H_A, H_B, H_C$ be the feet of the perpendiculars in $\triangle ABC$ from $A,B,C$, respectively. It is noted that $HH_B$ is perpendicular to $AD$, and as $H$ is the center of a circle $A$ and $D$ are on, $H_B$ is the midpoint of $AD$. Similarly, $H_C$ is the midpoint of $AE$. It is well known that $AH_A$ bisects $\angle H_CH_AH_B$. It is easily seen that $AX$ is perpendicular to $BC$ and thus passes through $H_A$, and so we note now that $\frac{AX}{AH_A} = \frac{AD}{AH_B} = \frac{AE}{AH_C} = 2$. Thus, $\angle DXE$ is the image of $\angle H_CH_AH_B$ after a homothety of magnitude $2$ about $A$, and so is bisected by the image of $AH_A$, $AX$, a segment containing $XH$, as desired.