$ CD $ is the polar of $ A $ and $ EF $ of $ B $. So $ N $ is the pole of $ AB $. So $ ON\perp AB $.

RSM wrote: $ CD $ is the polar of $ A $ and $ EF $ of $ B $. So $ N $ is the pole of $ AB $. So $ ON\perp AB $. Umm...........what are poles and polars?

Redeem wrote: RSM wrote: $ CD $ is the polar of $ A $ and $ EF $ of $ B $. So $ N $ is the pole of $ AB $. So $ ON\perp AB $. Umm...........what are poles and polars? See here:- http://www.math.ust.hk/excalibur/v11_n4.pdf

Draw the circle with radius OA; extend ON to meet this circle at point H. We do have ∠OHA = 90° and ON×NH = DN×NC because both D and C are also on this circle. Since CEDF is cyclic, EN×NF = DN×NC, or ON×NH = EN×NF. This implies that H is on the circle that has OB as its diameter, or ∠OHB = 90°. We then have ∠OHA + ∠OHA = 180°, or the three points A, H and B are collinear. In other words, ON is perpendicular to AB.

Toss the figure in the complex plane such that the circle with the center $O$ is the unit circle. Then it is easy to see that $$\bar{n}=\frac{c+d-e-f}{cd-ef}, \bar{a}=\frac{2}{c+d}, \bar{b}=\frac{2}{e+f}$$It suffices to show $$\frac{\bar{n}-\bar{o}}{\bar{a}-\bar{b}} \in iR$$Now, $$\frac{\bar{n}-\bar{o}}{\bar{a}-\bar{b}}=\left( \frac{c+d-e-f}{cd-ef} \right) \left( \frac{1}{\frac{2}{c+d}-\frac{2}{e+f}} \right)$$$$=\frac{(e+f)(c+d)}{2(ef-cd)}$$Conjugating this we see that $$\frac{n-o}{a-b}=\frac{(\frac{1}{e}+\frac{1}{f})(\frac{1}{c}+\frac{1}{d})}{2(\frac{1}{ef}-\frac{1}{cd})}=-\frac{(e+f)(c+d)}{2(ef-cd)}=-\frac{\bar{n}-\bar{o}}{\bar{a}-\bar{b}}$$proving the required result.

Challenge: Let $DE \cap CF=L$. Prove that $L$ lies on $AB$.

How this solves the problem

Fun Fact: If $K=AC \cap BE$, then $KE=KC$ and $K, L, N$ are collinear.

Proof

Attachments:

@above, Pascal on $DEECFF$, $EDDFCC$.

It's easy to see that points $B, E, O, F$ are concyclic and points $A, C, O, D$ are concyclic too. Now let's notice that $DN \cdot NC = EN \cdot NF$, so it means that $N$ has equal power to circumcircles of quadrilaterals $BEOF$ and $ACOD$, so $ON$ is radical axis of that circles. Let $M_{1}$ and $M_{2}$ be centres of that circles. It's clear that $M_{1}$ and $M_{2}$ are midpoints of $BO$ and $AO$ respectively. So $M_{1}M_{2}$ is parallel to $AB$, but also we have $M_{1}M_{2}$ is perpendicular to $ON$. So $AB$ is perpendicular to $ON$ as required.

Its equal to prove that the intersection of ON whith AB is the Miquel point of (CDEF)