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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.

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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.

Illustration

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