2. Let $M, N, R, S$ be the midpoints of line segments $BC, CA, BD, AD$. Let $Z$ be the centroid of $\triangle ABC$. Quadrilateral $QFMC$ is cyclic as $\angle QFC = 90^\circ = \angle QMC$. Note that therefore $CQ$ is a diameter of the circumcircle of $QFMC$. Analogously, we see that $PNEC$ is a cyclic quadrilateral, with diameter $CP$.
We show that $Z$ also lies on the circumcircles of these cyclic quadrilaterals. The similarity transforming triangle $BCA$ into triangle $BDC$ transforms triangle $CZM$ into $DFR$, as $C$ is mapped to $D$, the centroid $Z$ is mapped to the centroid $F$, and the midpoint $M$ of $BC$ is mapped to the midpoint of $BD$, which is $R$. So $\triangle CZM \sim \triangle DFR$, in particular $\angle CZM = \angle DFR = \angle CFM$ (opposite angles). Therefore $Z$ lies on the circumcircle of the cyclic quadrilateral $QFMC$. Analogously, we have $\angle CZN = \angle DES = \angle CEN$, from which follows that $Z$ lies on the circumcircle of the cyclic quadrilateral $PNEC$.
We can now show that $Z$ lies on $PQ$. As $CQ$ is a diameter of the circle through $Q, F, M, C, Z$, we have $\angle QZC = 90^\circ$. As $CP$ is a diameter of the circle through $P, N, E, Z, C$, we also have $\angle CZP = 90^\circ$. Hence $P, Z$, and $Q$ are collinear. $\square$