I just asked a question on math stack exchange and doing some progress on it: Proof of Pascal's Theorem (on circles) using complex numbers..
I did some complex number geometry, got the intersection points and now all it remains is to evaluate a determinant which should evaluate to zero.
Here's the determinant:
$$ \frac{i}{4} \begin{vmatrix} \frac{ab(d+e) - de(a+b)}{ab-de} & \frac{\frac{1}{ab}(\frac 1d+\frac1e) - \frac{1}{de}(\frac1a+\frac1b)}{\frac{1}{ab}-\frac{1}{de}} & 1 \\ \frac{bc(e+f) - ef(b+c)}{bc-ef} & \frac{\frac{1}{bc}(\frac1e+\frac1f) - \frac{1}{ef}(\frac1b+\frac1c)}{\frac{1}{bc}-\frac{1}{ef}} & 1 \\ \frac{cd(f+a) - fa(c+d)}{cd-fa} & \frac{\frac{1}{cd}(\frac1f+\frac1a) - \frac{1}{fa}(\frac1c+\frac1d)}{\frac{1}{cd}-\frac{1}{fa}} & 1 \\ \end{vmatrix} $$ where $a, b, c, d, e, f$ are points $A, B, C, D, E, F$ on the unit circle.
I don't want to evaluate this by straight up multiplication and minors. Please explain how to solve this determinant using its properties.
Thanks!
(This is only a comment. It is placed here so that the others can copy the matrices below easily.)
Your matrix is equal to $$ \frac{i}{4}\pmatrix{\frac{1}{ab-de}\\ &\frac{1}{bc-ef}\\ &&\frac{1}{cd-fa}} \pmatrix{ ab(d+e)-de(a+b) &(a+b)-(d+e) &ab-de\\ bc(e+f)-ef(b+c) &(b+c)-(e+f) &bc-ef\\ cd(f+a)-fa(c+d) &(c+d)-(f+a) &cd-fa}. $$ It suffices to prove that the rightmost matrix in the above is singular.