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Given the cyclic quadrilateral ABCD, let $AB \cap CD \equiv E, AD \cap BC \equiv F$. Let $M$ be an arbitrary point on $(ABCD)$. $P$ and $Q$ are the intersection of $ME, MF$ with $(ABCD)$.
Prove that $AC, BD$, and $ PQ$ are concurrent.

When I say duality I mean I used it to solve this, but other types of solutions are welcome too.

Mine involves duality as well. (Pascal is a result of dual projective axioms so far as I know )
Let $AC \cap BD=G, AC \cap PQ=G', AP \cap CQ=X, AQ \cap CP=Y$.
Applying Pascal's theorem on $ADCQMP$, we get $AD \cap MQ=F, DC \cap MP=E, AP \cap CQ=X$ are collinear.
Similarly applying Pascal's theorem on $AQMPCB$, we get $E,F,Y$ are collinear. So $EF, XY$ are the same line hence they have the same pole.
The pole of $EF$ is $AC \cap BD=G$ and the pole of $XY$ is $AC \cap PQ=G'$. So $G=G'$ from which the result follows.

Here \[\frac{AQ}{DQ}=\frac{\sin \angle ADQ}{sin \angle QAD}=\frac{\sin\angle AMF}{\sin\angle DMF}=\frac{AF\times DM}{DF\times AM}=\frac{AF\times \sin\angle DAM}{DF\times \sin\angle ADM}\]
Again
\[\frac{BE}{\sin\angle BME}=\frac{ME}{\sin\angle ABM}=\frac{ME}{\sin\angle ADM}\]
\[\Longrightarrow \sin\angle BME=\frac{BE\times \sin\angle ABM}{ME}\]
Similarly $\sin\angle CME=\dfrac{CE\times \sin\angle DAM}{ME}$
So \[\frac{BP}{CP}=\frac{\sin\angle BME}{\sin\angle CME}=\frac{BE\times \sin\angle ADM}{CE\times \sin\angle DAM}\]
So in cyclic hexagon $AQDCPB$
\[\frac{AQ\times CD\times BP}{DQ\times CP\times AB}=\frac{AQ}{DQ}\times \frac{BP}{CP}\times \frac{CD}{AB}\]
\[=\frac{AF\times \sin\angle DAM}{DF\times \sin\angle ADM}\times\frac{BE\times \sin\angle ADM}{CE\times \sin\angle DAM}\times \frac{CD}{AB}\]\[=\frac{AF\times BE\times CD}{DF\times CE\times AB}\]\[=\frac{AF}{DF}\times \frac{DC}{CE}\times \frac{EB}{BA}=1\] by using Manelau's theorem in $\triangle AED$.
So the three diagonals $AC,BD,PQ$ of hexagon $AQDCPB$ are concurrent.