Consider the Euler line and let $N_9$ be the nine point center and $O$ be the circumcenter. Since $S$ lies in $(ABC)$ and $H$ lies outside $(ABC)$, $(ABC)$ intersect segment $GS$ and thus intersects the circle with diameter $(GS)$ at two points one of which is $P$. Using the cross ratio the circle with diameter $HS$ is an apollionus circle with ratio $1:2$ and focii at $N_9$ and $O$. Thus $\frac{N_9P}{OP}=\frac{1}{2}$ as desired.

Notice that there exist homotheties at $H$ and $S$ with ratios $\frac12$ and $-\frac12$ that map $(DEF)$ to $(ABC)$. Thus by the Power Ratio Lemma, there is a circle through $H$ and $S$ coaxal with $(ABC)$ and $(DEF)$. But clearly, this must have center on the Euler Line and we are done (It's easy to check as the above commenter did that in the obtuse case, the circles really do intersect, but the coaxality holds regardless)

Let $X$ and $Y$ be the intersections of the circumcircle and the 9-point circle (these points obviously exist due to the obtuse triangle). Consider the homotheties: (1) in $H$ with factor $\frac{1}{2}$ (2) in $S$ with factor $-2$ Chaining these gives a resulting homothety (3) with factor $-1$ centered at some point $O$, i.e a reflection about the point $O$. Since (3) maps the circumcircle to the 9-point circle back to the circumcircle, $O$ must be the circumcenter of $ABC$. Finally: Let $X_1$ be the reflection of $H$ across $X$ and $X_2$ be the image of $X_1$ under (3). It follows that the angle $\angle HXS = 180 - \angle X_2 X X_1 = 90$ by Thales' theorem, placing it on the desired circle. This holds analagously for $Y$.

Alternatively just bash Let $X, Y$ be the intersections of the $HS$ circle and the circumcircle of $ABC$. These exist, since $H$ lies outside, and $S$ inside of $ABC$. We set $a, b, c$ on the unit circle and find the following relations: $o = 0$, $h = {a+b+c}$, $s = \frac{a+b+c}{3}$, $n = \frac{a+b+c}{2}$ (Here $N$ denotes the centre of the 9-point circle). We focus on $X$ The circle condition gives: 1) $| x | = 1$ 2) $\frac{x-h}{x-s} = - \overline{\left( \frac{x-h}{x-s} \right)}$ We reformulate (2) in terms of $n$: $ \frac{x - 2n}{x - \frac{2}{3}n} + \frac{\overline{x} - 2\overline{n} }{\overline{x} - \frac{2}{3}\overline{n}} = 0$ $ (x-2n)(3\overline{x}-2\overline{n}) + (\overline{x} - 2\overline{n})(3x - 2n) = 0 $ $ 6x\overline{x} + 8n\overline{n} = 8\overline{x}n + 8x\overline{n} $ $-\frac{1}{4} + x\overline{x} + n\overline{n} - \overline{x}n - x\overline{n} = 0 $ $ | x - n | ^ 2 = (x-n)\overline{(x-n)} = \left(\frac{1}{2}\right)^2$ Or in other words $X$ (and analagously $Y$) lie on the 9-point circle, as desired.

Yet another solution: Define $N$ as 9-point circle centre and $X$ as 9-point circle intersected with circumcircle The ratios of the Euler line are well-known and we easily find $-1 = (ON; SH)$ Note however that by angle-bisector theorem we have $\frac{NS}{SO} = \frac{1}{2} = \frac{XN}{XO}$ Now using the famous Lemma from projective geometry we get from $(ON;SH)$ harmonic and $XS$ being the angle bisector $\angle NXO$ that $\angle HXS=90$, placing $X$ on the Thales circle of $HS$, finishing the problem