Proving that four points are non-concyclic 
In the figure, G and H are the circum-center and the orthocenter of ⊿ABC respectively. AH produced meets BC at O. GR ┴ BC at R. BS is the diameter of the circumscribed circle. Show that B, O, H, and G are not concyclic. 
The derived facts are (1) AHCS is a parallelogram: and (2) AH = 2GR.
My questions are:
(1) Fact #1 helps, but does fact #2 also help?
(2) Does “if they are concyclic, then BH = HS ……” help?
(3) I can prove it, in a clumsy way, through angle-chasing. Would like to see if there is a more elegant proof. 
 A: Elaborating on my comment ...

Consider $B$ and $C$ fixed, with $\angle BGC \geq 90^\circ$, so that $\angle A \geq 45^\circ$. Let $T$ be the foot of the perpendicular from $C$ to $AB$. Note that $B$, $O$, $H$, $T$ are concyclic (since opposite angles at $O$ and $T$ are supplementary); call the circle $\gamma$.
For $A$ such that $\angle ABC \approx 90^\circ$, circle $\gamma$ is nearly a point-circle at $B$. Circumcenter $G$ lies outside $\gamma$.

For $A$ close to $S$, circle $\gamma$ pass through a point of $\overline{GS}$ (in fact, $\gamma$ contains $G$ itself, when $\angle A = 45^\circ$ and $A=S$). Circumcenter $G$ lies inside $\gamma$.

By continuity, we can position $A$ such that $G$ lies on $\gamma$.

Therefore, the assertion that $B$, $O$, $H$, $G$ are non-concyclic is false, in general.

We can be precise about this.
Let $G^\prime$ be the foot of the perpendicular from $H$ to $\overline{BS}$. Certainly, $B$, $O$, $H$, $G^\prime$ are concyclic; moreover, $B$, $O$, $H$, $G$ will be concyclic if and only if $G=G^\prime$.

Writing $d := |\overline{BS}|$ for the circumdiameter, we have 
$$\begin{align}
|BG| &= \frac{1}{2}d \\[6pt]
|BG^\prime| &= |BH| \;\cos\angle HBG^\prime \\
&= |BH|\;\cos(\angle CBS - \angle CBG) \\
&= |BH|\;\cos((90^\circ - A)-(90^\circ-C)) \\
&= |BH|\;\cos(C-A)
\end{align}$$
(With appropriate adjustment when $C-A > 90^\circ$.)
Now, as I pointed out recently in this answer, vertex-to-orthocenter lengths obey a Law-of-Sines-like relation, but with cosines; here, in particular, we have $|BH| = d\;|\cos B|$.
Therefore,
$$\begin{align}
G = G^\prime \qquad&\Leftrightarrow\qquad |\overline{BG}| = |\overline{BG^\prime}| \\
&\Leftrightarrow\qquad \frac{1}{2}d = d\;|\cos B|\;\cos(C-A) \\[6pt]
&\Leftrightarrow\qquad 1 = 2\;|\;\cos(C+A)\;\cos(C-A)\;| \\[6pt]
&\Leftrightarrow\qquad 1 = |\;\cos 2A + \cos 2C\;| \qquad (\star)
\end{align}$$
We can see that condition $(\star)$ holds in some special cases I've mentioned: $(1)$ equilateral $\triangle ABC$ (ie, $A=B=C=60^\circ$), and $(2)$ $A=S$ (ie, $C=90^\circ$) with $A=45^\circ$.
Thus, there are plenty of triangles for which the four points in question are concyclic; and there are plenty of triangles for which the points are non-concyclic.
A: 
To simplify the explanation, I have the angles labeled.
My proof is by the following contradiction.
If GBOH is a cyclic quadrilateral, then
k + h = p … [ext. angle of ‘cyclic quad’.]
= q … [angles in the same seg.]
= r … [alt. angles AS// HC]
∴ k + h + s = r + s
But h = r + s … [ext. angle of triangle]
∴ k + s = 0 which is impossible (unless TC on the other side of G)
[The ’unless’ part can be forced to not going to happen if A is located even closer to S.]
