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.
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.
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.
