Sorry, pretty new to this so formatting/explaining probably isn't that good: Let $R'$ be the point diametrically opposite $R$ wrt $\Phi$. Since the circles are orthogonal, inverting about $\Phi$ sends $S$ to $Q$, and that means that (S,Q;R,R') = -1. From power of a point, we can see that $AD$, $BC$, and $SQ$ all intersect at one point. Let this point be $X$. Now since $X=AD \cap BC$ and $R=AC\cap BD$, it follows that $X$ lies on the polar of $R$ wrt $\Gamma$, and hence as $RX$ intersects $\Gamma$ at $S$ and $Q$, we get that (S,Q;R,X) = -1. Therefore $X$ is diametrically opposite $R$, and so $\angle RAX = \angle CAD = 90^{\circ}$. So $C$ and $D$ are diametrically opposite wrt $\Phi$. By taking perspectivity at $A$, we can see that (S,Q;C,D) = (S,Q;R,R') = -1. As SCQD is a harmonic quad, and $CD$ is its diameter, it follows that it is a kite, with $CD$ perpendicularly bisecting $SQ$.