Equivalence of falling on one straight line in a six-point system

Question

Today I came up with a property that turns out to be a corollary of Pascal's theorem and its converse, but I wonder if the property would still work in 3D space

Let $A, B, C, D, E, F$ be six points in the plane, and let: $BD∩AC=M,AE∩FD=N,FC∩BE=O$

$EF∩BC=R,AB∩DE=Q,CD∩AF=P$

Points $M, N, O$ falling on one line is equivalent to points $R, Q, P$ falling on one line.

Assuming that points $M, N, O$ lie on one line, a conic section will pass through points $A, B, C, D, E, F$ according to the inverse of Pascal’s theorem, and then points $R, Q, P$ will lie on one line according to Pascal’s theorem.

Assuming that the points $R, Q, P$ lie on one line, a conic section will pass through the points $A, B, C, D, E, F$ according to the inverse of Pascal’s theorem, and then the points $M, N, O$ will lie on one line according to Pascal’s theorem.

Thus, the theorem was proven from both sides, so the equivalence is achieved.

My question clearly is, will the feature continue to work if points $A, B, C, D, E, F$ do not belong to the same plane?

Obviously, the same proof cannot be used in space

Answer 1

The same proof can also be used in space, because if points $PQR$ are aligned, then points $ABCDEF$ are forced to lie in the same plane. And the same if $MNO$ are aligned.

Proof. Suppose points $PQR$ are aligned. From the definition of point $Q$ it follows that $QAB$ are aligned, hence incident lines $PQR$ and $QAB$ define a plane $\alpha$.

But:

• $PAF$ are aligned, hence $F\in\alpha$;
• $REF$ are aligned, hence $E\in\alpha$;
• $RCB$ are aligned, hence $C\in\alpha$;
• $PDC$ are aligned, hence $D\in\alpha$.

In summary, points $PQRABCDEF$ all lie on $\alpha$. An analogous reasoning can be made when $MNO$ are aligned.