While answering a recent question I have found algebraically the following nice proposition related to Pascal's theorem.
Let $ABCD$ be a cyclic quadrilateral and let $E$ and $F$ be arbitrary diametrically opposite points on its circumcircle. Further let $G=(AE)\cap(DF),H=(BE)\cap(CF),I=(AC)\cap(BD)$, and $O$ be the center of the circumcircle.
Then: $$IG:IH=\tan\frac{\angle AOD}2:\tan\frac{\angle BOC}2.\tag1$$
The proposition reveals that the ratio $IG:IH$ does not depend on the choice of the antipode points $E,F$, i.e. it is an "internal" property of cyclic quadrilateral.
My algebraic proof is very lengthy and cannot be reproduced here (essentially it is the same computation as reported in the linked answer).
Is this proposition known and can it be easily proved?
The collinearity of the points $G,I,H$ can be assumed as it follows from Pascal's theorem.
I present the Menelaus's theorem proof here... hope it is simpler. Let $EA\cap DB=Y$ and $EB\cap DF=X$, then, using the Menelaus's theorem to triangle $EGH$ and line $YIB$, then use that theorem to triangle $XEG$ with line $DYB$, we have
$$\frac{GI}{HI}=\frac{GY}{YE}\frac{EB}{BH}=\frac{EB}{BH}\frac{GD}{DX}\frac{XB}{EB}=\frac{GD}{BH}\frac{XB}{DX}$$
Notice that $\frac{XB}{DX}=\frac{FB}{DE}$ (notice that if $EA\parallel DB$, this also holds.) Also notice that $\angle EDG=\angle EDF=90^\circ$, now we have:
$$\frac{GI}{HI}=\frac{GD}{BH}\frac{FB}{DE}=\frac{GD/DE}{BH/FB}=\frac{\tan \angle DEA}{\tan \angle CFB}=\frac{\tan \frac{\angle AOD}2}{\tan \frac{\angle BOC}2}$$