Let $H \ne A$ be the intersection of $(ABEE')$ and $(ADFF'),$ if $E'$ and $F'$ are the other points on $CD$ and $CB$ satisfying the angle conditions. Angle chase utilizing the weird angle condition to get $\angle DHB = 90^\circ$ and is collinear with $AC.$ Power of point at $C$ implies $(EE'FF')$ cyclic, done. $\blacksquare$

Wow this seemed really hard for a p7 (at least synthetically.)

Let $E',F'$ denote reflections of $E,F$ respectively across perpendicular bisectors of $AB,AD;O=AC\cap BD,G$ denotes the intersection of circle with diameter $BD$ with ray $OC.$ Since $2\angle AFD=\angle AOD=2\angle AGD,2\angle AEB=\angle AOB=$ $=2\angle AGB$ we get $G\in \odot (ABEE')\cap \odot (BCFF').$ By radical axis $EE'FF'$ are concyclic with center $K,$ done.

Denote the $\omega_B = \odot(ABE)$ and $\omega_D=\odot(ADF)$. Let $\omega_B$ meet $CD$ again at $E'$, and let $\omega_D$ meet $CB$ again at $F'$. Notice that since $AB$ and $EE'$ share the same perpendicular bisector, we have $KE=KE'$. Similarly, we have $KF=KF'$, so it suffices to show that $E,E',F,F'$ are concyclic. Let $M$ be the midpoint of $BC$. If we let $X$ be the point on ray $MC$ such that $MX=MB=MD$, then by an easy angle chasing, we have $X\in\omega_B\cap\omega_D$. By power of point at $C$, we are done. khina wrote: Wow this seemed really hard for a p7 (at least synthetically.) I found this problem at a normal IMO 1/4-level. The first thing I noticed after sketching the diagram is that there are two choices of $E$ that satisfies the condition. Normally, getting the coordinates of $E$ will involve solving quadratic, which makes it hard to do anything synthetically. Thus, the first thing I did was add another intersection $E'$ (and $F'$) as well. Then, I have this solution in a few minutes. Having seen the solution of IMO 2012 P5 really helps with this.

@above what is $M$? edit: ah, this solution is very smart. I maintain that I believe this is pretty tricky to do synthetically, but that solution does make it more reasonable.

Edited, thanks for pointing out.

Let lines $BC, CD$ meet $(ABD)$ at $B'$ and $D'$. Let the angle bisectors of $\angle CAD'$ and $\angle CAB'$ meet $CD, BC$ at $X,Y$ respectively. By angles, we have that $ADXF$ and $ABYE$ are cyclic. Let $E'$ and $F'$ denote the other intersections of $(ADX)$ and $(ABY)$ with $BC, CD$. By power of point and angle bisectors, $$CE.CE' = CB.CY = (BC) \left(\frac{CB'.AC}{AC + AB'} \right) = \left(\frac{BC.CB'.AC}{AC + BD} \right)$$and similarly, $CF.CF' = \left( \frac{CD.CD'.AC}{AC + BD} \right)$ and since $CD.CD' = CB.CB'$ by PoP, we have that $CE.CE' = CF.CF'$ implying that the points $E,F,E',F'$ are concyclic. Observe that $EE'BA$ is an isosceles trapezoid and since $K$ lies on the perpendicular bisector of $AB$, we have that $KE = KE'$. Analogously, we have $KF = KF'$. So since $EFE'F'$ is cyclic, it follows that $K$ is the center of the circle, and hence $KE = KF$, as desired. $\blacksquare$

Let $P$ be the intersection point of the diagonals of the parallelogram and let the circumcircle of $\triangle ABD$ meet lines $BC$ and $CD$ again at $Q,R$, respectively. Denote $\Gamma_1$ and $\Gamma_2$ the circumcircle of $\triangle AEB$ and $\triangle AFD$, respectively. Suppose that $\Gamma_1$ meets line $CD$ again at $E_1$ and $\Gamma_2$ meets line $BD$ again at $F_1$. Clearly, $BF\cdot FQ=BF\cdot BF_1$ and $DE\cdot ER=DE\cdot DE_1$. Extend lines $BD$ to meets $\Gamma_1$ and $\Gamma_2$ again at $X,Y$ respectively. Then we have $$\angle AYX+\angle AXY=\angle AYD+\angle AXB=\angle AFD+\angle AEB=90^{\circ}$$by the given condition, i.e. $\angle XAY=90^{\circ}$. Moreover, since $\angle APY=\angle APB=2\angle AEB=2\angle AXY$, we obtain that $P$ is the midpoint of $XY$ and hence $BY=DX$. By power of point, we have $$BF\cdot FQ=BF\cdot BF_1=BY\cdot BD=DX\cdot DB=DE\cdot DE_1=DE\cdot ER$$ and hence $KE=KF$ by power of point with respect to the circumcircle of $\triangle ABD$.

Let $\omega_B,\omega_D$ be circles $\odot(ABE),\odot(ADF)$. Let $E',F'$ be the other intersections of $\omega_B,\omega_D$ with $CD,CB$ respectively. Note that $AB,EE'$ share the same perpendicular bisector. We want to show that $\square EFE'F'$ are cyclic because it implies that $K$, the intersection of perpendicular bisectors of $EE',FF'$, is its center. That's why $KE = KF$. Let $O_B,O_D$ be the centers of $\omega_B,\omega_D$. By angle chasing for the problem's statement, $O_B,A,B,M$ are concyclic. Moreover, $O_D,A,D,M$ are concyclic. We perform $\sqrt{BC}$ inversion on $\triangle ABD$. It suffices to show that Quote: Let $ABD$ be a triangle and $M$ be a point on its circumcircle such that $\square ABMD$ is a harmonic quadrilateral. Let $C$ be the midpoint of $AM$. Points $O_B,O_D$ lie on $\overrightarrow{BM},\overrightarrow{DM}$ such that $BO_B = BA$ and $DO_D = DA$. Let $\omega_B,\omega_D$ be the perpendicular bisectors of $AO_B,AO_D$. Let $E,E'$ be intersections of $\omega_B$ with $\odot(ACD)$ and $F,F'$ be intersections of $\omega_D$ with $\odot(ACD)$. Prove that $E,E',F,F'$ are concyclic. Obviously, $\omega_B,\omega_D$ are internal bisectors of $\angle ABM$ and $ADM$ which intersect at a point called $X$ because $\square ADMB$ is harmonic. Therefore, $XE \cdot XE' = XA \cdot XC = XF \cdot XF'$. We get that $E,E',F,F'$ are concyclic as desired.

Let $M$ be the midpoint of $BC$, the projections of $K$ onto $BC, CD$ be $N, P$ respectively, the external and internal bisectors of $\angle AMB$ meet $(ABM)$ at $X$ and $X_1$ respectively, the external and internal bisectors of $\angle AMD$ meet $(ADM)$ at $Y$ and $Y_1$ respectively, the projection of $X_1$ onto $AM$ be $Q$, and the projection of $Y_1$ onto $AM$ be $Q_1$. It's clear that $X, K, X_1$ and $Y, K, Y_1$ are collinear along the perpendicular bisectors of $AB$ and $AD$ respectively. Now, observe that $XA = XB$ and $$\angle AXB = \angle AMB = \angle ADM + \angle DAM$$$$= \angle ADB + \angle ACB = 2 \angle AEB$$which means $X$ is the circumcenter of $(ABE)$. Similarly, we deduce that $Y$ is the circumcenter of $(ADF)$. Because $M \in BD$, we know $Y \in MX_1$, $Y_1 \in MX$, and $\angle X_1MY_1 = 90^{\circ}$. Now, we claim that $Q = Q_1$. Since $X_1$ is the midpoint of arc $AB$ (not containing $M)$ and $Y_1$ is the midpoint of arc $AD$ (not containing $M$), we have $$MQ = \frac{MA + MB}{2} = \frac{MA + MD}{2} = MQ_1$$which implies $Q = Q_1$, as $Q$ and $Q_1$ both lie on segment $MA$. It follows that $AM \perp X_1Y_1$. Now, Pythag and the Perpendicularity Lemma yield $$MX^2 - MY^2 = (XX_1^2 - MX_1^2) - (YY_1^2 - MY_1^2)$$$$= ((AX^2 + AX_1^2) - MX_1^2) - ((AY^2 + AY_1^2) - MY_1^2)$$$$= (AX^2 - AY^2) + (AX_1^2 - AY_1^2) - (MX_1^2 - MY_1^2)$$$$= AX^2 - AY^2$$so the Converse of the Perpendicularity Lemma gives $AM \perp XY$. Since $C \in AM$ by the parallelogram condition, we can apply the Perpendicularity Lemma once again, yielding $$KE^2 - KF^2 = (KP^2 + PE^2) - (KN^2 + NF^2)$$$$= (KP^2 - KN^2) + (PE^2 - NF^2)$$$$= ((CK^2 - CP^2) - (CK^2 - CN^2))$$$$+ (XE^2 - XP^2) - (YF^2 - YN^2)$$$$= (CN^2 - CP^2) + (AX^2 - XP^2) - (AY^2 - YN^2)$$$$= (CN^2 + YN^2) - (CP^2 + XP^2) + (AX^2 - AY^2)$$$$= (CY^2 - CX^2) + (CX^2 - CY^2) = 0$$as desired. $\blacksquare$ Remarks: My solution absolutely sucks. It's far more natural to consider $E'$ and $F'$, as the condition that $E$ and $F$ must lie on segments $CD$ and $BC$ respectively is unnecessary. By the way, $MQ = \frac{MA + MB}{2}$ is a result from IMO 1987/2. (It's likely that this is also a lemma from somewhere else.)

[asy][asy] //setup size(9cm); pen darkgrn=RGB(13,83,76); pen thickgrn=darkgrn+linewidth(1); //actual code pair A,B,C,D,M,K; A=(0,0); B=(3,4); D=(7,0); C=B+D; M=C/2; K=circumcenter(A,B,D); draw(A--B--C--D--A); draw (A--C^^B--D,darkgrn); draw(circumcircle(A,M,B),gray); draw(circumcircle(A,M,D),gray); pair Ob,Od; Ob=extension(K,B/2,M,incenter(A,M,D)); Od=extension(K,D/2,M,incenter(A,M,B)); draw(circumcircle(Ob,Od,M)); draw(Ob--Od,thickgrn); draw(Ob--M--Od,darkgrn); path wb,wd; wb=circle(Ob,distance(Ob,A)); wd=circle(Od,distance(Od,A)); draw(wb,darkgrn); draw(wd,darkgrn); pair E,F; E=intersectionpoints(wb,C--D)[0]; F=intersectionpoints(wd,C--B)[0]; //labels label("$A$",A,dir(-90)); label("$B$",B,dir(90)); label("$C$",C,dir(90)); label("$D$",D,dir(-90)); label("$M$",M,dir(60)); label("$O_b$",Ob,dir(-60)); label("$O_d$",Od,dir(90)); label("$E$",E,dir(-30));label("$F$",F,dir(70)); [/asy][/asy] As earlier we define $\omega_b,\omega_d=(ABE),(ADF)$ and $O_b,O_d$ as their respective centers. Thus we have $O_b,O_d$ as the intersections of the bisectors of $\angle AMD,\angle AMB$ with the $(AMB),(AMD)$ respectively. Claim: $\overline{AC}\perp\overline{O_bO_d}$ and thus $C$ lies on the radical axis of $\omega_b,\omega_d$. Proof: Invert about $M$ with arbitrary radius to obtain the following diagram, ignoring everything to do with sides $CB,CD$. First, we show that $\overline{O_b^*O_D^*}\parallel\overline{B^*D^*}$; this follows by angle bisector theorem: \[\frac{A^*O_b^*}{D^*O_b^*}=\frac{A^*M}{D^*M}=\frac{A^*M}{B^*M}=\frac{A^*O_d^*}{B^*O_d^*}.\][asy][asy] pen darkgrn=RGB(13,83,76); pen thickgrn=darkgrn+linewidth(1); pair A,B,D,M; A=(-0.4,1); B=(-2,0); D=-B; M=(0,0); pair Ob,Od; Ob=extension(A,D,M,incenter(A,M,B)); Od=extension(A,B,M,incenter(A,M,D)); draw(A--B--D--A--M); draw(Ob--M--Od,darkgrn); draw(Ob--Od, thickgrn); label("$A^*$",A,dir(90)); label("$B^*$",B,dir(-90)); label("$D^*$",D,dir(-90)); label("$M$",M,dir(-90)); label("$O_b^*$",Ob,dir(60)); label("$O_d^*$",Od,dir(100)); draw(Ob--A--Od,dotted); [/asy][/asy] It follows that $\overline{AM}$ bisects $\overline{O_b^*O_d^*}$; the result then follows by isogonality.$\qquad\qquad\square$ Now we finish by one of the many isomorphic PoP solutions; use the fact that $C$ has the same power with both $\omega_b,\omega_d$.

Define $M$ as the intersection point of $AC$ and $BD$. Let $\omega _1$ and $\omega _2$ be the circumcircle of triangle $AMB$ and $AMD$ respectively . $O_1$ and $O_2$ be midpoint of arc$AMB $ and arc$AMD$ respectively. Again $EC$ and $FC$ intersects $(AEB)$ and $(AFD)$ at $G$ and $H$ respectively. Now observe, $$\angle AO_1B= \angle AMB= \angle ADB+\angle ACB=2\angle AEB$$That implies $O_1$ is the center of $(AEB)$. Simmilarly, $O_2$ is the center of $(AFD)$. Define $N$ as the intersection of $(AEB)$ and $(AFD)$. Let $MO_2$ and $MO_1$ meet $\omega_1$ and $\omega_2$ at $X$ and $Y$ respectively. Observe, $X$ is the midpoint of arc$AB$(of $\omega_1$) not containing $M$. So, $XO_1$ is diameter and $O_1B\perp XB$ and $XB$ is tangent to $AEBN$. That implies $$\angle BNM=\angle XBA=\angle AMX\Rightarrow XM \parallel BN$$ Simmilarly $MY \parallel ND$ and $\angle BND=\angle XMY=\angle O_2MY=90°$. So, $\angle BMN+\angle BMA=180°-2\angle BNM+2\angle AXM=180°$ that implies $A,M,N$ are collinear and $C$ lies on the radical axis of $(AEB)$ and $(AFD)$. So we can say $F,E,H,G$ lies on a circle. Since $ABGE$ is cyclic, so perpendicular bisector of $AB$ and $EG$ goes through $K$. Simmilarly, perpendicular bisector of $FH$ goes through $K$. So, $K$ is the center of $EHGF$ which gives $KE=KF$.

My solution basicaly uses the bellow perpendicular criterion and then trig bash. Lemma: Given four distinct points $A,C,B,D$ in the plane, we have $AC \perp BD $ if, and only if, $${AB}^2-{AD}^2={CB}^2-{CD}^2.$$

Let $M$ and $N$ be the circumcenters of $(ABE)$ and $(ADF),$ respectively. Then by problems condition, $$\angle AMB = \angle ADB + \angle ACB = \angle ADB + \angle DAC = \angle AXB$$so that point $M$ lies on the circumcircle of $(AXB).$ Furthermore, we have that $M$ also lies on the perpendicular bissector of $AB;$ hence $M$ must be the midpoint of arc $AXB$ in $(AXB);$ indeed, it cannot be the other arc-midpoint since the agnle $\angle ANE$ is acute and so $M$ must be on the same semiplane that $E$ with respect to $AB$ and hence it must be on the same semiplane that $X$. Similarly, $N$ is the midpoint of the arc $AXD$ in $(AXD).$ Now notice that both $K$ and $M$ lie on the perpendicular bissector of $AB,$ in particular we have $KM \perp AB \Rightarrow KM \perp ED$ and by the Lemma we have $${KE}^2 = {ME}^2-{MD}^2+{KD}^2 $$similarly, we have ${KF}^2= {NF}^2 -{NB}^2 + {KB}^2;$ and then it is sufficient to show $$ {ME}^2-{MD}^2+{KD}^2 = {NF}^2 -{NB}^2 + {KB}^2$$But by construction $KD =KB, ME = MB$ and $NF= ND,$ so it reduces to $${MB}^2 -{MD}^2 = {ND}^2 - {NB}^2 $$Now I present to ways to finish from above.

You're welcome for an unreasonably long solution. I'm glad I finally got around to solving this, albeit totally in the wrong way. I think this proof is still instructive, though. There are some nice ideas, maybe? Let $(ABE)$ and $(ADF)$ intersect lines $CD$ and $CB$ at $E'$ and $F'$, respectively. Note that the circumcenters of $ABEE'$ and $ADFF'$ lie on the perpendicular bisectors of $EE'$ and $FF'$ respectively. Since $K$ lies on both of these perpendicular bisectors, we have that $KE=KE'$ and $KF=KF'$. Thus, it suffices to show that $EE'F'F$ is a cyclic quadrilateral, since that would force it to have center $K$. Let $ABEE'$ have center $O_1$ and let $ADFF'$ have center $O_2$. Let $P$ be the intersection of the diagonals of $ABCD$. Since $\angle AO_1B = \angle AEB = \angle ADB + \angle ACB = \angle APB$, we have that $O_1$ is the midpoint of arc $APB$ (in the circumcircle of $(APB)$). Similarly, $O_2$ is the midpoint of arc $APD$. Now, note that $EE'F'F$ is cyclic if and only if $CE\cdot CE'=CF\cdot CF'$, which is equivalent to $C$ lying on the radical axis of 1. the circle with center $O_1$ and radius $O_1A$ and 2. the circle with center $O_2$ and radius $O_2A$. Since $A$ obviously lies on this radical axis, it suffices to show that $P$ does as well, i.e. that $AP\perp O_1O_2$. Thus, we reduce the problem to the following statement (which has been relabeled): Let $ABC$ be a triangle. Let $M$ be the midpoint of $BC$, and let $P$ and $Q$ be the midpoints of arcs $AMB$ and $AMC$, respectively. show that $PQ\perp AM$. Proof. It is clear that $MQ$ and $MP$ are the external angle bisectors of angles $AMC$ and $AMB$, respectively, so they are the internal angle bisectors of $\angle AMB$ and $\angle AMC$, respectively. Let $R$ and $S$ be diametrically opposite to $P$ and $Q$ in their respective circles. Finally, let $D$ and $E$ be the intersections of $MP$ and $MQ$ with $AC$ and $AB$, respectively. Claim. We have that $RS\parallel PQ$. Proof. We'll show that $QM\cdot MS = PM\cdot MR$. First, we have $\angle QMP = \angle QMA + \angle AMP = \frac 12 (\angle AMB + \angle AMC) = 90^\circ$. Let $O$ be the circumcenter of $\triangle ABC$. Let $X$ and $Y$ be the feet from $M$ to lines $PR$ and $QS$, respectively. It is clear that $OXMY$ is cyclic with diameter $OM$. Thus, $\angle XYM= \angle XOM = \angle ABC$ and $\angle YXM = \angle YOM =\angle ACB$, so $\triangle ABC\sim\triangle MYX$. In particular, $\frac{MX}{MY} = \frac{AC}{AB}$. Since $ARBP$ and $AQCS$ are cyclic kites with $\angle APB = \angle AMB = \angle ASC$, they are similar. In particular, $\frac{RP}{QS} = \frac{AB}{AC}$. Combining these two, we have that $[MPR] = [MQS]$, implying the desired equality $QM\cdot MS = PM\cdot MR$. Now, we claim that $DERS$ is a cyclic quadrilateral. Proof. We use this well-known lemma: Let $ABC$ be a triangle with $M$ being the midpoint of minor arc $BC$. If $AM$ hits $BC$ at $D$, then $MD\cdot MA = MB^2 = MC^2$. (The proof is by similar triangles.) Applying this in $AMBR$, we have that $RE\cdot RM = RA^2$. By power of a point, we have that $RE\cdot EM = \frac 14 AB^2$. Now, by Stewart's theorem in $\triangle AMB$ with cevian $ME$, we have that \[MB^2\cdot EA + MA^2 \cdot EB = ME^2\cdot AB + AB\cdot \frac{AB^2}{4}.\]Using the Angle Bisector theorem, we can rewrite $AE = \frac{AM}{AM+MB}\cdot AB$ and $EB=\frac{MB}{AM+MB}$. In particular, we have \[ \frac{AM\cdot MB^2}{AM+MB}\cdot AB + \frac{MB\cdot MA^2}{AM+MB}\cdot AB=ME^2\cdot AB + AB\cdot \frac{AB^2}{4},\]which reduces to $AM\cdot MB = ME^2 + AB^2/4$. This gives \[AM\cdot MB = ME^2 + AB^2/4 = ME^2 + ME\cdot RE = ME\cdot MR.\]Symmetrically, we have $AM\cdot MC = MD\cdot MS$, but since $MB=MC$, this gives that $ME\cdot MR = MD\cdot MS$, as desired. Since $DERS$ is cyclic and $RS\parallel PQ$, we have that $DPEQ$ is cyclic by Reim's theorem. Also, since (by Angle Bisector theorem) \[\frac{AE}{EB} = \frac{AM}{MB} = \frac{AM}{MC} = \frac{AD}{DC},\]we have that $DE\parallel BC$. Now we're ready to finish. We have \[\measuredangle(AM, PQ) =\measuredangle(AM,BC) + \measuredangle(BC,PM) + \measuredangle(PM,PQ) = \angle AMB + \angle CMP - \angle MPQ.\] Since $\angle MPQ = \angle DPQ = \angle DEQ = \angle DEM =\angle EMB$, \[\measuredangle(AM, PQ) =\angle AMB + \angle CMP + \angle EMB = \angle AMB + (90^\circ - \frac 12 \angle AMB) - \frac{1}{2}\angle AMB = 90^\circ\] so indeed $AM\perp PQ$, as desired. This completes the proof.

PROA200 wrote: You're welcome for an unreasonably long solution. I'm glad I finally got around to solving this, albeit totally in the wrong way. I think this proof is still instructive, though. There are some nice ideas, maybe? My solution has a setup comparable as yours, where $P \equiv X$, $Q \equiv Y$, $R \equiv X_1$, and $S \equiv Y_1$. I think it's pretty interesting that both of us managed to produce two distinct freaky solutions while starting with similar observations.

Let $O=AC\cap BD$, $X=CD\cap \odot(ABE)$, and $Y=BC\cap \odot(ADF)$, and $Z=\odot(ADF)\cap \odot(ABE)$. Claim 1. $\overline{A-Z-C}$ are collinear. Proof. $\angle BZD=\angle BZA+\angle AZD=\angle BEA+\angle AFD=\frac{\angle DOC}{2}+\frac{\angle BOC}{2}=90^{\circ}\implies \angle ODZ=\angle OZD.$ $$\angle DZC=180^\circ-\angle DCZ-\angle ZDC=180^\circ-\angle DCZ-\angle ODC+\angle ODZ=\angle DOC+\angle ODZ=180^\circ-\angle DZA.\square$$ By PoP we have $CE\cdot CX=CZ\cdot CA=CF\cdot CY\implies EXYF$ is cyclic. But we can also notice that $K$ is on the perpendicular bisecors of $EX$, and $FY\implies K$ is the center of $EXYF\implies KE=KF.\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13.461366295343204, xmax = 12.347049478215052, ymin = -5.851963976503904, ymax = 10.42904597410537; /* image dimensions */ pen xfqqff = rgb(0.4980392156862745,0,1); pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures */ draw((-7.166752246469837,5.039291766000371)--(-5,-2), linewidth(0.4)); draw((-5,-2)--(5,-2), linewidth(0.4)); draw((-7.166752246469837,5.039291766000371)--(2.833247753530163,5.03929176600037), linewidth(0.4)); draw((5,-2)--(2.833247753530163,5.03929176600037), linewidth(0.4)); draw((-7.166752246469837,5.039291766000371)--(5,-2), linewidth(0.4)); draw((-5,-2)--(2.833247753530163,5.03929176600037), linewidth(0.4)); draw(circle((-2.166752246469837,2.725215117187501), 5.509532714902509), linewidth(0.4) + xfqqff); draw((-7.166752246469837,5.039291766000371)--(2.9907793950259585,-2), linewidth(0.4) + qqwuqq); draw((2.9907793950259585,-2)--(2.833247753530163,5.03929176600037), linewidth(0.4) + qqwuqq); draw((-7.166752246469837,5.039291766000371)--(4.337053609073509,0.15376403949812456), linewidth(0.4) + red); draw((4.337053609073509,0.15376403949812456)--(-5,-2), linewidth(0.4) + red); draw(circle((-1.2389072207213987,3.010813492811954), 6.265306916192372), linewidth(0.4) + red); draw(circle((-2.1667522464698377,1.405988665326745), 6.180686970019157), linewidth(0.4) + qqwuqq); draw((2.833247753530163,5.03929176600037)--(1.7652731980763523,8.508901411843453), linewidth(0.4)); draw((-5,-2)--(-7.324283887965634,-2), linewidth(0.4)); draw((2.833247753530163,5.03929176600037)--(3.4767257584496756,-1.1143381843803044), linewidth(0.4)); draw((-5,-2)--(3.4767257584496756,-1.1143381843803044), linewidth(0.4)); /* dots and labels */ dot((-7.166752246469837,5.039291766000371),linewidth(4pt) + dotstyle); label("$A$", (-7.5,5.203042780082639), NE * labelscalefactor); dot((-5,-2),linewidth(4pt) + dotstyle); label("$B$", (-5.2,-2.4), NE * labelscalefactor); dot((5,-2),linewidth(4pt) + dotstyle); label("$C$", (5.151245080291488,-2.2741617898268047), NE * labelscalefactor); dot((2.833247753530163,5.03929176600037),linewidth(4pt) + dotstyle); label("$D$", (2.92014371668949,5.203042780082639), NE * labelscalefactor); dot((-2.166752246469837,2.725215117187501),linewidth(4pt) + dotstyle); label("$K$", (-2.064659329916778,3.0121414410500336), NE * labelscalefactor); dot((-1.0833761232349184,1.5196458830001855),linewidth(4pt) + dotstyle); label("$O$", (-1.602359047368616,1.404140458273809), NE * labelscalefactor); dot((4.337053609073509,0.15376403949812456),linewidth(4pt) + dotstyle); label("$E$", (4.688944797743327,0.01723961062931543), NE * labelscalefactor); dot((2.9907793950259585,-2),linewidth(4pt) + dotstyle); label("$F$", (3.0608438026824087,-2.434961888104427), NE * labelscalefactor); dot((1.7652731980763523,8.508901411843453),linewidth(4pt) + dotstyle); label("$X$", (1.854843065600247,8.660244893051523), NE * labelscalefactor); dot((-7.324283887965634,-2),linewidth(4pt) + dotstyle); label("$Y$", (-7.250462499370072,-1.8319615195633427), NE * labelscalefactor); dot((3.4767257584496756,-1.1143381843803044),linewidth(4pt) + dotstyle); label("$Z$", (3.7241442080775973,-1.1485611018834474), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

Here is a solution which revolves around properly characterizing $E$ and $F$. Switch $C$ and $D$. We first provide an alternate characterization of $E$ (and similarly $F$). Define $M$ to be the midpoint of $\overline{BC}$. Let the internal angle bisector of $\angle AMC$ intersect the perpendicular bisector of $\overline{AB}$ at $X$, so $ABMX$ is cyclic. Then draw the circle centered at $X$ through $A$ and $B$. Then $E$ is one of its intersection points with $\overline{CD}$; let the other intersection point be $E'$. We can define $Y,F,F'$ similarly. Since $\overline{AB} \parallel \overline{CD}$, it follows that $\overline{AB}$ and $\overline{EE'}$ share a perpendicular bisector, so $KE=KE'$. Likewise, $KF=KF'$, so it suffices to show that $EE'FF'$ is cyclic. By radical center it suffices to show that $(ABEE')$ and $(ACFF')$ intersect again at some point on the $A$-median $\overline{AMD}$. This is equivalent to $\overline{XY} \perp \overline{AM}$, since the second intersection of $(ABEE')$ with $\overline{AM}$ is the reflection of $A$ over the foot from $X$ to $\overline{AM}$. Since $\angle XMY=90^\circ$, it suffices to prove that $(XMY)$ is tangent to $\overline{BC}$, since this means $\measuredangle EMA=\measuredangle BME=\measuredangle MDE$ and triangle similarity finishes. This is equivalent to showing the midpoint of $\overline{XY}$ lies on the perpendicular bisector of $\overline{BC}$. Suppose the $(ABEE')$, which is centered at $X$, intersects $\overline{BC}$ again at $P$. Because $X$ lies on the $\angle AMC$-bisector, it follows that $MA=MP$. Likewise if $(ACFF')$ intersects $\overline{BC}$ again at $Q$, $MA=MQ$. Thus the midpoint of the projections from $X$ and $Y$ onto $\overline{BC}$ is just $\tfrac{B+P+C+Q}{4}=M$, done. $\blacksquare$ Remark: It is also possible to coordbash the "midpoint of $\overline{XY}$" claim very cleanly (which is how I first did it) with the following setup. Let $M=(0,0),C=(1,0),B=(-1,0),A=(a,b)$. By the angle bisector theorem, if $\overline{MX}$ and $\overline{MY}$ intersect $\overline{AC}$ and $\overline{AB}$ again at $R$ and $S$ respectively, then $AR/AC=2AM/BC=AS/AB \implies \overline{RS} \parallel \overline{BC}$. Then we can parametrize $R=(at-(t-1),bt)$ and $S=(at+(t-1),bt)$ for some $t$. Since $\angle RMS=90^\circ$ by considering the slopes we get the relation $(at)^2+(bt)^2=(t-1)^2$. On the other hand, if we compute the $x$-coordinates of $X$ and $Y$, their numerators are miraculously the same, and the problem boils down to proving the denominators sum to zero, or $$\frac{bt}{at+(t-1)}+\frac{bt}{at-(t-1)}+\frac{2a}{b}=0 \iff \frac{2abt^2}{(at)^2-(t-1)^2}+\frac{2a}{b}=0 \iff \frac{2abt^2}{-(bt)^2}+\frac{2a}{b}=0,$$which is true.

Theoretically, two $E$ on $CD$ would work. So we construct $E$ and $E'$. Analogously, $F$ and $F'$. Notice that $KE=KE'$ and $KF=KF'$. Thus, it suffices to show $E, E', F, F'$ are concyclic. It suffices to show that $C$ lies on the radical axis of $(ABE)$ and $(ADF)$. Since $A, M, C$ are collinear, it suffices to show that $M$ lies on the radical axis of $(ABE)$ and $(ADF)$. Let $P=AC\cap BD$. Angle chasing gives us the circumcenter ($O_1$) of $(ABE)$ lies on $(ABP)$. Similar for $O_2$ . Now, we can ignore $E$ and $F$, and the stupid condition. It suffices to show $O_1O_2\perp AM$. Now take $X$ and $Y$ as the circumcenter of $(ABP)$ and $(ADP)$. By Simon's Theorem, we conclude $XY\parallel O_1O_2$. We're done.