Angle chasing, parallelogram, cyclic quadrilateral In parallelogram  $ ABCD $ with $ \angle BAD > 90^{\circ},$ show that the circle passing through the projections of $C$ onto $ AB,BD$ and $DA$, respectively, passes through the center of the parallelogram.
 A: Let $P$, $Q$ and $R$ be  projections of $C$ on $AD$, $AB$ and $BD$  respectively.
Also, let $S$ be a center of $ABCD$.
Thus, $S$ is a center of circumcircle of $PAQC.$
Thus,
$$\measuredangle PSQ=360^{\circ}-2\measuredangle PAQ=360^{\circ}-2\left(180^{\circ}-\measuredangle ADC\right)=2\measuredangle ADC.$$
In another hand, since $PAQC$ and $RQBC$ are cyclics, we obtain:
$$\measuredangle PQR=\measuredangle PQC-\measuredangle RQC=\measuredangle PAC-\measuredangle RBC=\measuredangle DAC-\measuredangle DBC.$$
Also, since $PAQC$ and $DPRC$ are cyclics, we obtain:
$$\measuredangle QPR=\measuredangle QPC-\measuredangle RPC=\measuredangle QAC-\measuredangle RDC=\measuredangle BAC-\measuredangle BDC.$$
Thus, $$\measuredangle PRQ=180^{\circ}-\measuredangle PQR-\measuredangle QPR=180^{\circ}-\measuredangle DAC+\measuredangle DBC-\measuredangle BAC+\measuredangle BDC=$$
$$=180^{\circ}-\measuredangle DAB+180^{\circ}-\measuredangle DAB=2\measuredangle ADC,$$
which says that $$\measuredangle PSQ=\measuredangle PRQ$$ and we are done! 
A: Since $$ \angle DCP = 90^{\circ}-\angle PDC = 90^{\circ}-\angle QBC = \angle QCB =: x$$ 
so we have $$\angle  DRP = x \;\;\;{\rm and} \;\;\; \angle QRB = x$$
since $CRPD$ and $BCRQ$ are both cyclic. So $$  \angle PQR = 180^{\circ}-2x$$
On the other hand $AQCR$ is also cyclic with center at $S$ so $$\angle PSQ = 2\angle PCQ = 2(90^{\circ}-x)$$ and thus $PQRS$ is cyclic.
