<s>[INST] Collinearity in geometry Let S be the intersection of diagonals in a cyclic quadrilateral. Let p be a circumcircle of a triangle ABS and it intersects BC in M and q is a circumcircle of a triangle ADS and q intersects CD in N. Prove that M, N and S are collinear.
I tried proving that vectors NC and DC are the same, and also that vectors CM and CB are equal but nothing seemed to work. I'm probably not thinking right.
 [/INST] Brute force computation
You tagged your question as linear algebra, so here you get a linear algebra proof. With a bit of projective geometry thrown in, due to my background.
W.l.o.g. assume $A,B,C,D$ on the unit circle, with the following homogeneous coordinates (using a rational parametrization to avoid square roots and trigonometric functions):
$$
A=\begin{pmatrix}1\\0\\1\end{pmatrix}\qquad
B=\begin{pmatrix}b^2-1\\2b\\b^2+1\end{pmatrix}\qquad
C=\begin{pmatrix}c^2-1\\2c\\c^2+1\end{pmatrix}\qquad
D=\begin{pmatrix}d^2-1\\2d\\d^2+1\end{pmatrix}
$$
Then you get
$$
S=\begin{pmatrix}bc - bd + cd - 1\\2c\\bc - bd + cd + 1\end{pmatrix} \\
p: (bc-bd+cd+1)(x^2+y^2)+2b(d-c)xz+2(d-c)yz+(bc-bd-cd-1)z^2=0 \\
q: (bc-bd+cd+1)(x^2+y^2)+2d(b-c)xz+2(b-c)yz+(cd-bd-bc-1)z^2=0 \\[2ex]
M=\begin{pmatrix}
b c^{3} -  b c^{2} d + c^{3} d + b c + 3 c^{2} -  b d - 3 c d - 1 \\
2 c^{2} d + 4 c - 2 d \\
b c^{3} -  b c^{2} d + c^{3} d + b c + c^{2} -  b d + c d + 1
\end{pmatrix} \\[1ex]
N=\begin{pmatrix}
b c^{3} -  b c^{2} d + c^{3} d - 3 b c + 3 c^{2} -  b d + c d - 1 \\
2 b c^{2} - 2 b + 4 c \\
b c^{3} -  b c^{2} d + c^{3} d + b c + c^{2} -  b d + c d + 1
\end{pmatrix} \\
\det(M,N,S)=0
$$
which proves the collinearity. Obviously not a proof you'd want to attempt without help from a computer algebra system.
Angle chasing
If the above is not the kind of thing you want to compute yourself, I suggest you turn your known cocircularities into angle equalities, and follow them though. A bit like what I did in the second half of this post.
You have to show that $\angle NSD=\angle MSB$ to show the collineariry. Let's follow that backwards to something which only uses $A,B,C,D$:
\begin{align*}
\angle NSD&=180°-\angle DNS-\angle SDN \\
&=180°-\angle DNA-\angle ANS-\angle BDC \\
&=180°-\angle DSA-\angle ADS-\angle BDC \\
&=\angle SAD-\angle BDC \\
&=\angle CAD-\angle BDC \\
\angle MSB&=180°-\angle SBM-\angle BMS \\
&=\angle CBD-\angle BMA+\angle SMA \\
&=\angle CBD-180°+\angle ASB+\angle SBA \\
&=\angle CBD-\angle BAS \\
&=\angle CBD-\angle BAC \\
&=\angle CAD-\angle BDC=\angle NSD \tag*{$\Box$}
\end{align*}
Note that the computation seems less symmetric than it actually is, due to the fact that I tried to make all angles positive in the following angle. For the proof to be universal, you'll have to consider signed angles, though.

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