Not so hard as an intermediate P5, but very interesting. Let ME cross NF at G, AD and BC at H, AB and CD at J. The projections of O, P on AD, BC are O1, O2, P1, P2. Obviously EQFH concyclic. Easy angle chasing gets MN//EF, so GE=GF. Furthermore, angle MEF=90+ADB, so ABO~EFG. Since AO1/O1D=BO2/O2C, AP1/P1D=BP2/P2C, P1E/O1E=P2F/O2F=PQ/OQ, so AE/ED=BF/FC. Construct I, the similarity center of AD and BC. ABCD concyclic gets OQPI, HIJ collinear and HIO=90. IAB~IEF~ICD, so IABO~IEFG, so IOG~IAE. Then angle IOG=IAD=180-IJD, so OG is perpendicular to CD.

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Because APD ~ BPC, AE : ED = BF : FC so if R = BC \cap AD and S = Miquel Point of complete quadrilateral determined by ABCD then QESRF is cyclic, furthermore by angle chasing <CRM = <DRN = <OBA = <OAB. If M1, N1 = (RCD) \cap RM, RN and M2, N2 = (RAB) \cap RM, RN then spiral similarity at S takes M1, M, M2 to C, F, B to D, E, A to N1, N, N2 so ME \cap NF is collinear with CN1 \cap DM1 and AM2 \cap BN2; the former is clearly on the perpendicular bisector of CD since <DCN1 = <CDM1 and the latter is O, so we are done.

v4913 wrote: Because $\triangle APD \sim \triangle BPC, AE:ED = BF:FC$ so if $R = BC \cap AD$ and $S$ = Miquel Point of complete quadrilateral determined by $ABCD$ then $QESRF$ is cyclic, furthermore by angle chasing $\angle CRM = \angle DRN = \angle OBA = \angle OAB$. If $M_1, N_1 = (RCD) \cap RM, RN$ and $M_2, N_2 = (RAB) \cap RM, RN$ then spiral similarity at $S$ takes $M_1, M, M_2$ to $C, F, B$ to $D, E, A$ to $N_1, N, N_2$ so $ME \cap NF$ is collinear with $CN_1 \cap DM_1$ and $AM_2 \cap BN_2$; the former is clearly on the perpendicular bisector of $CD$ since $\angle DCN_1 = \angle CDM_1$ and the latter is $O$, so we are done. LTFY (Latexed that for you)

LoloChen wrote: Not so hard as an intermediate P5, but very interesting. Let ME cross NF at G, AD and BC at H, AB and CD at J. The projections of O, P on AD, BC are O1, O2, P1, P2. Obviously EQFH concyclic. Easy angle chasing gets MN//EF, so GE=GF. Furthermore, angle MEF=90+ADB, so ABO~EFG. Since AO1/O1D=BO2/O2C, AP1/P1D=BP2/P2C, P1E/O1E=P2F/O2F=PQ/OQ, so AE/ED=BF/FC. Construct I, the similarity center of AD and BC. ABCD concyclic gets OQPI, HIJ collinear and HIO=90. IAB~IEF~ICD, so IABO~IEFG, so IOG~IAE. Then angle IOG=IAD=180-IJD, so OG is perpendicular to CD. splendid solution(and diagrame)

Outline of solution (I found this outline but couldn't solve it, apparently there is a solution using this outline so yay): 1. Prove that the intersection is linear w.r.t. $Q$ (I proved this during the test) 2. Prove two special cases

LLL2019 wrote: Outline of solution (I found this outline but couldn't solve it, apparently there is a solution using this outline so yay): 1. Prove that the intersection is linear w.r.t. $Q$ (I proved this during the test) 2. Prove two special cases I solved it exactly like this, call $T$ the intersection and move $Q$ linearly. Clearly $M$, $N$ move linearly too, but $\triangle MNT$ has a fixed shape $\implies$ $T$ moves linearly, so we just need to check two cases. With some $MMP$, isn't hard to prove: $P=Q \implies T $ is the midpoint of segment $CD$ We only need one more case, so just consider $O = Q$ and complex bash with $(ABC)$ the unit circle.