$F$ is a point outside $\square ABCD$ such that $\widehat{CFD} = 135^\circ$. $FF' \perp CD$ at $F'$. $AC \cap BD = \{E\}$. $AF \cap BD = \{M\}, AF \cap CD = \{M'\}$ and $BF \cap AC = {N}, BF \cap DC = \{N'\}$. $MC \cap ND = \{E'\}$.
Prove that $EF'$ passes through the circumcenter of $\triangle E'M'N'$.
When I first saw the problem, I couldn't believe my eyes. It's the hardest problem I have ever encountered.
First, since $\angle CFD = {3\pi \over 4}$, $BCFD$ is cyclic.
Of course, $ABCFD$ is inscribed in a circle with centre $E$.
Now suppose that $DG$ intersects $\bigcirc E$ at M (other than $D$).
So $\angle CDG = \angle CBG(\text{symmetric}) = \angle CDF$.
Similarly, $\angle KCD = \angle FCD$.
Therefore, $\triangle CKD \simeq \triangle CFD$, and these two triangles are symmetric about $CD$.
Second, consider that
$${CI \over ID} = {CF \cdot \sin \angle CFI \over DF \cdot \sin \angle IFD} = {CF \cdot 1 \over DF \cdot {1 \over \sqrt 2}} = {CF \over DF} \cdot \sqrt 2$$
and $${CG \over GE} = {CD \cdot \sin \angle CDG \over DE \cdot \sin \angle EDG} = \sqrt 2 \cdot {\sin \angle CDK \over \sin \angle KCL} = {CK \over KD} \cdot \sqrt 2 \:(\angle EDG = \angle DBF = \angle DCF)$$
$\implies {CI\over ID} = {CG\over GE}$, therefore $\triangle CGI$ is an isosceles right triangle, and so is $\triangle HJD$.
Denote the intersect of $CI$ and $HJ$ as O.
It is clear that $\angle HOI = {\pi \over 2}$, $\angle HKI = \angle HFI = {\pi \over 4}$, and $HO=OI$.
So $O$ is the circumcentre of $\triangle KHI$.
The last important thing to note is that$${LH \over HC} = {KL \cdot \sin \angle LKH \over KC \cdot \sin \angle CKH} = KL \cdot \sqrt 2 \cdot {\sin \angle LDK \over CK}$$
$$(\angle HKD = \angle HFD = {\pi \over 2} = \angle KLD,\text{ so }\angle HKL = \angle KDL)$$
$${LI \over ID} = {KL \cdot \sin \angle LKI \over KD \cdot \sin \angle DKI} = KL \cdot \sqrt 2 \cdot {\sin\angle KCL \over KD}$$
And ${KC \over \sin \angle KDC} = {KD \over \sin \angle KCD} \implies {HL \over HC} = {LI \over ID}$.
Now since $\triangle OHI \sim \triangle ECD$. It is easy to show that $\triangle HLO \sim \triangle CLE$.
$\implies \angle CLO = \angle CLE$. Now $L,O,E$ are collinear.