Well, let $X, Y$ be the intersections of $AM$ with $BC, PQ$. Since $AS$ is diameter, we have $AB \perp BS$ and $AC \perp CS$. This implies that $AB = BP, AC = CQ$. Thus we get $XM := d, MY = 2d, AX = 3d$ by looking at $\triangle{APQ}$. Since $G$ is centroid we also get $AG = 2GX = 2d$ meaning that $G$ is midpoint of $AM$. The conclusion follows!

Same solution as @above. Also note that by homothety $\circ (ABC)$ and $\circ (APQ)$ are tangent at $A$

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Note that we don't even need to know that points $A,O,S$ are collinear, as this can independently be deduced from $S \in \omega$. Indeed, note that if $O$ is the circumcenter of $\omega$ and $N$ the midpoint of $BC$, then by Brocard's theorem $ON \perp PQ,$ and so $BC \parallel PQ$. Hence, $\angle SAC=\angle QAC=90^\circ-\angle ASQ=90^\circ-\angle APQ=90^\circ-\angle ABC=90^\circ-\angle ASC,$ and so points $A,O,S$ are collinear. Now, for the problem at hand, note that as $A,O,S$ are collinear, we must have $AB=BP,AC=CQ$. Moreover, from Pascal's theorem the tangents to $\omega$ at $B,C$ meet on $X \in PQ,$ and so $h_a=XN,$ that is $2h_a=a\tan \angle A,$ which, since $ah_a=bc\sin \angle A,$ rewrites as $a^2=2bc\cos \angle A$. Due to the Law of Cosines this implies that $b^2+c^2=2a^2$. Now, to finish, note that $NG \cdot NA=AN^2/3=(2b^2+2c^2-a^2)/12=a^2/4=BN^2,$ and so $G$ is the $A-$ humpty point, hence $BGCM$ is a parallelogram, implying that $GN=NM,$ i.e. $AG=2GN=GM$. Thus, $\angle AGO=90^\circ,$ as desired.

My English is not good, so I only can use some easy words It is easy to know AC=CQ, so we have M is the barycentric of triangle APQ with Newton‘s theorem And then it's very simple

Easy question Just observe B and C are midpoints And then M is centroid We are done by parallel lines