I think that there is a mistake in the problem statement: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15.27058258054524cm); 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 = -25.378140919274287, xmax = 239.89244166127096, ymin = -73.24186140875749, ymax = 67.35154735893092; /* image dimensions */ pair A = (3.337520485388558,10.002028637562127), B = (-5.892316005167974,-13.054801208974956), C = (55.79921393750017,-12.535421340653524), D = (46.1824832155051,10.362739599369766), F = (71.01905019497728,45.276431532725006), G = (88.78260884036051,5.774685492280314), K = (74.0224384337187,22.88210207760651); /* draw figures */ draw(circle(D, 42.84648111272434), linewidth(2.)); draw(circle((66.5246594009278,-28.66810114776507), 74.0809925820767), linewidth(2.)); draw(A--B, linewidth(2.)); draw(B--C, linewidth(2.)); draw(C--D, linewidth(2.)); draw((55.31852510478256,44.56041799663973)--C, linewidth(2.)); draw(A--C, linewidth(2.)); draw(B--F, linewidth(2.)); draw(B--G, linewidth(2.)); draw(B--K, linewidth(2.)); draw((55.31852510478256,44.56041799663973)--G, linewidth(2.)); draw(A--D, linewidth(2.)); /* dots and labels */ dot(A,dotstyle); label("$A$", (3.9576646837507123,11.488683497851616), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (-5.2487849469858565,-11.449419819407202), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (56.38761512319965,-10.981295261912125), NE * labelscalefactor); dot(D,linewidth(4.pt) + dotstyle); label("$D$", (46.86908245413303,11.644725017016642), NE * labelscalefactor); dot((55.31852510478256,44.56041799663973),linewidth(4.pt) + dotstyle); label("$E$", (55.91949056570457,45.817817714157336), NE * labelscalefactor); dot(F,linewidth(4.pt) + dotstyle); label("$F$", (71.67968400137225,46.59802530998246), NE * labelscalefactor); dot(G,linewidth(4.pt) + dotstyle); label("$G$", (89.46841718618529,6.9634794420658634), NE * labelscalefactor); dot(K,linewidth(4.pt) + dotstyle); label("$K$", (74.64447286550777,24.12804655021872), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

You have a problem with diagram anyway since $AB\parallel CD$, not $AD\parallel BC$

My mistake.

No problem, since I can't label points on my diagram correctly ;p

this was the easiest question of the lot

I solved this problem.The main claim is that angle FDB is 90 degrees.

My solution during the test: Let. E' the symmetrical point of E to BD. 1 Claim) ACE' and ADF is similar this quite obvious if you find out <ACE=<ADF. 2 From ACE' and ADF we can find that line BC and BD are equivalent, since <ADB= <ACB. So, since G is a point which is symmetric to F by BD and E is a point which is symmetric to E' by BC, we can find out E and G are equivalent. So, G, D, F are collinear points, and BD is vertical to GF. 3 Let) N is intersection point of EC and GF. then, N, A,B,C,D are cyclic. Also, as ADG and ACE are similar, we can know that A,G,E,N are cyclic too. So, <GEA= <GNA= <ABD Finally, we can get A, B,C,E are cyclic

OR you can set F' that satisfies that angle F'BD is 90 degrees and AF'=AD Than by using angle chasing and similarities the problem is solved.

Anyway, it's a easy problem!

Let $O$ be the circumcenter of $BAE$. $ABCD$ is an isosceles trapezoid $\implies DO=CO$. We have $BC=AD=FD$ and $BO=FO$ so $$\triangle BCO\equiv\triangle FDO\ (s,s,s).$$$CE=AC=BD$ and $BO=EO$ thus $$\triangle BOD\equiv\triangle EOC\ (s,s,s).$$Hence $$\square BCEO\equiv FDBO\implies BE=FB$$and $90^\circ=\angle BCE=\angle BDF$. We have $\triangle BGD\equiv\triangle BFD\ (s,s,s)\implies \angle GBD=\frac{\angle GBF}{2}$ Let $X$ be the intersection of lines $EG, BD$. We want to prove $\angle EXB=\angle EFB$. $$\angle EXB=\angle EFB\iff 180^\circ=\angle EFB +\angle EGB+\angle GBD\iff 180^\circ=90^\circ-\frac{\angle EBF}{2} +90^\circ-\frac{\angle EBG}{2}+\angle GBD\iff \angle GBD=\frac{\angle GBF}{2}$$QED [asy][asy] import graph; size(12cm); real labelscalefactor =1.0; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -99.26695545175397, xmax = 74.50171442697476, ymin = -18.218012732399515, ymax = 115.87437416318622; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen duqqff = rgb(0.8313725490196079,0,1); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen qqffff = rgb(0,1,1); draw((35.19242198479023,55.9758978851566)--(35.91879273580862,53.21682219749131)--(38.6778684234739,53.943192948509704)--(37.95149767245552,56.70226863617499)--cycle, linewidth(1.6)); /* draw figures */ draw((-49.01675238067838,-0.2155584970685218)--(52.827785299256995,0.19558323305531267), linewidth(0.5) + wrwrwr); draw((-49.01675238067838,-0.2155584970685218)--(-34.59717198608807,56.40939298234919), linewidth(0.5) + dtsfsf); draw((37.95149767245552,56.70226863617499)--(52.827785299256995,0.19558323305531267), linewidth(0.5) + dtsfsf); draw((-49.01675238067838,-0.2155584970685218)--(37.95149767245552,56.70226863617499), linewidth(0.5) + wvvxds); draw(circle((-34.59717198608807,56.40939298234919), 58.432092456136324), linewidth(1.2) + dtsfsf); draw(circle((1.7415820337595618,40.598456013141885), 65.13211414477179), linewidth(1.2) + duqqff); draw(circle((52.827785299256995,0.19558323305531267), 119.23684410314071), linewidth(1.2) + sexdts); draw((-2.9947806073894796,105.55812936437658)--(52.827785299256995,0.19558323305531267), linewidth(0.5) + sexdts); draw((52.827785299256995,0.19558323305531267)--(-62.56165732110106,30.240574886237344), linewidth(0.5) +linetype("4 4") + sexdts); draw((-34.59717198608807,56.40939298234919)--(-2.9947806073894796,105.55812936437658), linewidth(0.5) + dtsfsf); draw((-49.01675238067838,-0.2155584970685218)--(1.7415820337595618,40.598456013141885), linewidth(0.5) + duqqff); draw((1.7415820337595618,40.598456013141885)--(-2.9947806073894796,105.55812936437658), linewidth(0.5) + duqqff); draw((-34.59717198608807,56.40939298234919)--(1.7415820337595618,40.598456013141885), linewidth(0.5) + qqffff); draw((1.7415820337595618,40.598456013141885)--(52.827785299256995,0.19558323305531267), linewidth(0.5) + duqqff); draw((-62.56165732110106,30.240574886237344)--(1.7415820337595618,40.598456013141885), linewidth(0.5) + duqqff); draw((1.7415820337595618,40.598456013141885)--(37.95149767245552,56.70226863617499), linewidth(0.5) + qqffff); draw((-66.1995633647867,7.260656600321844)--(52.827785299256995,0.19558323305531267), linewidth(0.5) + sexdts); draw((-66.1995633647867,7.260656600321844)--(-34.59717198608807,56.40939298234919), linewidth(0.5) + dtsfsf); draw((-66.1995633647867,7.260656600321844)--(-56.218100692155936,70.3115418723857), linewidth(0.5) + wrwrwr); draw((-62.56165732110106,30.240574886237344)--(37.95149767245552,56.70226863617499), linewidth(0.5) + wvvxds); draw((-56.218100692155936,70.3115418723857)--(52.827785299256995,0.19558323305531267), linewidth(0.5) + wvvxds); draw((-34.59717198608807,56.40939298234919)--(37.95149767245552,56.70226863617499), linewidth(0.5) + wrwrwr); /* dots and labels */ dot((-49.01675238067838,-0.2155584970685218),dotstyle); label("$A$", (-56.98293649938276,-7.8269541588511964), NE * labelscalefactor); dot((52.827785299256995,0.19558323305531267),dotstyle); label("$B$", (52.847878629393236,-3.6924582843320533), NE * labelscalefactor); dot((-34.59717198608807,56.40939298234919),dotstyle); label("$D$", (-39.91939855323909,56.57913161803123), NE * labelscalefactor); dot((37.95149767245552,56.70226863617499),dotstyle); label("C", (36.842869561615586,53.44463574351209), NE * labelscalefactor); dot((-62.56165732110106,30.240574886237344),dotstyle); label("E", (-74.44996206908384,29.393526847377164), NE * labelscalefactor); dot((-2.9947806073894796,105.55812936437658),dotstyle); label("$F$", (-7.371396919607237,107.26663819396106), NE * labelscalefactor); dot((1.7415820337595618,40.598456013141885),dotstyle); label("$O$", (0.9324710650052394,36.92529582044918), NE * labelscalefactor); dot((-66.1995633647867,7.260656600321844),dotstyle); label("G", (-76.46740018687095,0.242798312297393), NE * labelscalefactor); dot((-56.218100692155936,70.3115418723857),dotstyle); label("X", (-63.45617659408857,67.62523144638811), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

We doesn't need the condition $EC \perp BC$ to solve this problem..

Let the line $BD$ intersect $\Omega$ at $H$. Lemma: $\angle BDF=90^\circ$. Proof: Let $F'$ lie on the line perpendicular to $BD$ at $D$ such that $DF'=DA$ and $F',B$ lie on different sides of $CD$. It suffices to show that $ABF'E$ is cyclic. Let $F'_*$ be the reflection of $F'$ across the perpendicular bisector of $AB$. Then \begin{align*} \begin{cases} CE=CA\\CB=CF'_*\\ \angle BCA = \angle F'_*CE \end{cases}\Rightarrow \Delta CF'_*E \cong \Delta CBA \end{align*}and thus $ABF'_*E$ is an isosceles trapezium, hence cyclic. Since $ABF'F'_*$ is cyclic, $ABF'E$ is cyclic, so $ F=F'$. $\square$ Now, we know $BFG$ is an isosceles triangle with $D$ as the midpoint of $GF$. Also, \begin{align*} BF^2 = FD^2 + BD^2 = AD^2 + AC^2 = BC^2 + CE^2 = BE^2 \end{align*}Therefore, \begin{align*} \angle BHG + \angle BHE = \angle BHF + \angle BHE = \angle BEF + \angle BHE = \angle BFE + \angle BHE = 180^\circ. \ \ \ \square \end{align*}

[asy][asy] size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -22.37504457590086, xmax = 10.710287729459797, ymin = -18.88800127211882, ymax = 9.487675189062376; /* image dimensions */pen qqwuqq = rgb(0,0.39215686274509803,0); pen zzttqq = rgb(0.6,0.2,0); draw((-3.748751569966619,-2.0037915557363783)--(-4.218826306342471,-2.4147996332135637)--(-3.8078182288652855,-2.8848743695894155)--(-3.337743492489434,-2.47386629211223)--cycle, linewidth(1) + qqwuqq); draw(arc((-9.193495918045437,-6.247704277561425),0.8830604707836474,-57.980406554994985,19.707978842978694)--(-9.193495918045437,-6.247704277561425)--cycle, linewidth(1) + qqwuqq); draw(arc((-6.774309592056426,-10.116267262334594),0.8830604707836474,44.33120804703133,122.01959344500503)--(-6.774309592056426,-10.116267262334594)--cycle, linewidth(1) + qqwuqq); draw((3.7046746612428523,-10.528362148700294)--(0.8788811547351855,-2.6396886096997147)--(-9.193495918045437,-6.247704277561425)--cycle, linewidth(1) + zzttqq); draw((2.97052791898932,3.04174708678487)--(-3.337743492489434,-2.47386629211223)--(3.7046746612428523,-10.528362148700294)--cycle, linewidth(1) + zzttqq); /* draw figures */draw((-6.774309592056426,-10.116267262334594)--(3.7046746612428523,-10.528362148700294), linewidth(1)); draw((-3.337743492489434,-2.47386629211223)--(-6.774309592056426,-10.116267262334594), linewidth(1)); draw((0.8788811547351855,-2.6396886096997147)--(3.7046746612428523,-10.528362148700294), linewidth(1)); draw((-9.193495918045437,-6.247704277561425)--(0.8788811547351855,-2.6396886096997147), linewidth(1)); draw((-9.193495918045437,-6.247704277561425)--(3.7046746612428523,-10.528362148700294), linewidth(1)); draw((-3.337743492489434,-2.47386629211223)--(0.8788811547351855,-2.6396886096997147), linewidth(1)); draw(circle((-1.2859292065188226,-3.993441836652048), 8.222609710410591), linewidth(1)); draw(circle((-3.337743492489434,-2.47386629211223), 8.379515448183673), linewidth(1)); draw((-3.337743492489434,-2.47386629211223)--(3.7046746612428523,-10.528362148700294), linewidth(1)); draw((-9.646014903968185,-7.989479671009328)--(2.97052791898932,3.04174708678487), linewidth(1)); draw((-7.096240380383733,1.824770475785459)--(-9.646014903968185,-7.989479671009328), linewidth(1)); draw((-7.096240380383733,1.824770475785459)--(-3.337743492489434,-2.47386629211223), linewidth(1)); draw((-9.193495918045437,-6.247704277561425)--(-6.774309592056426,-10.116267262334594), linewidth(1)); draw((-6.774309592056426,-10.116267262334594)--(0.8788811547351855,-2.6396886096997147), linewidth(1)); draw((3.7046746612428523,-10.528362148700294)--(0.8788811547351855,-2.6396886096997147), linewidth(1) + zzttqq); draw((0.8788811547351855,-2.6396886096997147)--(-9.193495918045437,-6.247704277561425), linewidth(1) + zzttqq); draw((-9.193495918045437,-6.247704277561425)--(3.7046746612428523,-10.528362148700294), linewidth(1) + zzttqq); draw((2.97052791898932,3.04174708678487)--(-3.337743492489434,-2.47386629211223), linewidth(1) + zzttqq); draw((-3.337743492489434,-2.47386629211223)--(3.7046746612428523,-10.528362148700294), linewidth(1) + zzttqq); draw((3.7046746612428523,-10.528362148700294)--(2.97052791898932,3.04174708678487), linewidth(1) + zzttqq); /* dots and labels */dot((-6.774309592056426,-10.116267262334594),dotstyle); label("$A$", (-6.332779356664598,-9.851349121099497), NE * labelscalefactor); dot((3.7046746612428523,-10.528362148700294),dotstyle); label("$B$", (3.822416057347347,-10.234008658439077), NE * labelscalefactor); dot((0.8788811547351855,-2.6396886096997147),dotstyle); label("$C$", (0.878881154735189,-2.3747704684646176), NE * labelscalefactor); dot((-3.337743492489434,-2.47386629211223),linewidth(4pt) + dotstyle); label("$D$", (-3.448115152104683,-2.05098162917728), NE * labelscalefactor); dot((-9.193495918045437,-6.247704277561425),linewidth(4pt) + dotstyle); label("$E$", (-9.070266816093906,-6.024753747703692), NE * labelscalefactor); dot((2.97052791898932,3.04174708678487),linewidth(4pt) + dotstyle); label("$F$", (3.0865323316943076,3.2768165445507247), NE * labelscalefactor); dot((-9.646014903968185,-7.989479671009328),linewidth(4pt) + dotstyle); label("$G$", (-10.424292871295497,-8.232404924662811), NE * labelscalefactor); label("$r$", (-3.036020265738981,-2.816300703856441), NE * labelscalefactor); dot((-7.096240380383733,1.824770475785459),linewidth(4pt) + dotstyle); label("$K$", (-8.187206345310258,1.657872348114038), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] CLAIM. $F,D,G$ are collinear. Proof. Let $F'$ be the point on $\Gamma$ such that $F'D\perp BD$ and $F'$ lies on the same side of $BD$ as $F$. Then from $DF=DA=BC$ and $DB=AC=CE$ we have $\triangle BDF\cong\triangle ECB$, hence $BE=BF$. Therefore, \begin{align*} \angle EF'B&=180^{\circ}-\angle EAB\\ \iff 90^{\circ}-\frac{1}{2}\angle EBF&=180^{\circ}-\angle EAB\\ \iff 2\angle EAB-\angle EBF&=180^{\circ}\\ \iff 2\angle EAB-\angle EBD-\angle CEB&=180^{\circ}\\ \iff \angle EAB+\angle CAB+\angle AEB-\angle EBD&=180^{\circ}\\ \iff \angle CAB-\angle EBD&=\angle EBA\\ \iff \angle CAB&=\angle ABD \end{align*}which is true. Hence $F'$ lies on $\Omega$, which implies $F=F'$. Therefore, since $DF=DG$ we justify our claim. $\blacksquare$ Now assume $GE$ meet $BD$ at $K$, then $KG=KF$, hence $B$ is the point on the angle bisector of $\angle EKF$ such that $BE=BF$, by the converse of Fact $5$ we have $B\in (KEF)$, hence $K$ lies on $\Omega$ as deisred.

First, we pinpoint the location of $F$. Note that $K$, the center of $\Omega$ lies on both the perpendicular bisectors of $AE$ and $AB$, hence\[\angle FDB = \angle FDK + \angle BDK = \angle ADK + \angle ACK = \angle BCK + \angle ECK = 90^{\circ}.\]Thus, $G$ is the reflection of $F$ over $D$, hence $\angle GDA = \angle GDB - \angle ADB = 90^{\circ} - \angle ACB = \angle ECA$. Hence $AGD$ and $AEC$ are spirally similar, hence so are $AGE$ and $ADC$, thus $\angle GEA = \angle DCA = \angle DBA$, so if we let $P' = BD \cap \Omega$ again, then $\angle P'EA = \angle 180^{\circ} - P'BA = 180^{\circ} - \angle GEA$ thus $G, E, P'$ collinear as desired. $\blacksquare$ Remark: The very last step probably requires directed angles in order to work for all configurations, but I'm too lazy to edit it. Also, I think this solution is new

We use complex numbers. Set $(ABE)$ as the unit circle, and WLOG $b=\overline a=\frac1a$. $c$ satisfies that $c+a+e-ae\overline c=a+e \implies d=\overline c=\frac c{ae}$. Now note that $f$ satisfies: $|f|^2=f\overline f=1$, $\left|\frac{f-d}{a-d}\right|^2=\frac{f-d}{a-d}\times\overline{\left(\frac{f-d}{a-d}\right)}=1$. Expanding, $f\overline f-d\overline f-\overline d f+d\overline d=a\overline a-d\overline a-\overline d a+d\overline d$. Since $a\overline a=f\overline f=1$, this simplifies to $$\overline d f+\frac df=\overline d a+d\overline a\iff\overline d f^2-(\overline d a+d\overline a)f+d=0$$This is a quadratic equation, and clearly one of the solutions is $a$. By Vieta's, the other solution $f=\frac d{\overline d a}=\frac 1{a^2e}$. Let $BD$ intersect $\Omega$ again at $P$. Note that G is the reflection of F about BP, so $g=b+p-bp\overline f$. Finally, we wish to show $P,G,E$ collinear, which is equivalent to $$\frac{g-p}{\overline g-\overline p}=\frac{p-e}{\overline p-\overline e}=-pe$$Note $$\frac{g-p}{\overline g-\overline p}=\frac{b-bp\overline f}{\frac1b-\frac{f}{bp}}=b^2\left(\frac{1-p\overline f}{1-\frac f p}\right)=-\frac{b^2p}f=-pe$$and so we are done. Fun fact: Note that the above proof doesn't use any properties of $p$. So the above proof generalises the problem to hold for any point $G$ such that $BF=BG$! yet another victory for complex numbers