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$F$ is a point outside $square ABCD$ such that $widehatCFD = 135^circ$. $FF' perp CD$ 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$angle CFD=3pi over 4$, $BCFD$ 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 $angle CDG=angle CBG(textsymmetric)=angle CDF$.

Similarly, $angle KCD=angle FCD$.

Therefore,$triangle CKD simeq triangle CFD$, and these two triangles are symmetric about $CD$.

Second, consider that

$$CI over ID=CF*sinangle CFI over DF*sinangle IFD=CF*1 over DF * 1 over sqrt 2=CF over DF*sqrt 2$$.

And$$CG over GE=CD*sin angle CDG over DE*sin angle EDG=sqrt 2*sinangle CDK over sin angle KCL=CK over KD*sqrt 2 $$.($angle EDG=angle DBF=angle DCF$)

So $CIover ID=CGover GE$. Therefore $triangle CGI$ is an isosceles right triangle. So is $triangle HJD$.

Denote the intersect of $CI$ and $HJ$ as O.

It is clear that $angle HOI=pi over 2$, and $angle HKI =angle HFI =pi over 4$,and $HO=OI$.

So $O$ is the circumcentre of $triangle KHI$.

Last step.

Note that$$LH over HC=KL*sinangle LKH over KC*sin angle CKH=KL*sqrt 2 *sinangle LDK over CK$$
($angle HKD=angle HFD=pi over 2=angle KLD$,so$angle HKL=angle KDL$).

$$LI over ID=KL*sinangle LKI over KD*sinangle DKI=KL*sqrt 2 *sinangle KCL over KD$$.

And $KC over sinangle KDC=KD over sinangle KCD$.

So $HL over HC=LI over ID$.

Now since$triangle OHI sim triangle ECD$. It is easy to show $triangle HLO sim triangle CLE$.

So $angle CLO=angle CLE$. Now $L,O,E$ are collinear.