Hello. My solution. [asy][asy]import graph; size(8cm); 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 = -6.348523849398338, xmax = 5.854266843587791, ymin = -2.97752245227218, ymax = 3.926688071391023; /* image dimensions */ pen qqqqcc = rgb(0.,0.,0.8); draw(arc((-0.7454801878199087,0.5695050580200022),0.2293757649057543,-108.87222294123652,-37.37779865153979)--(-0.7454801878199087,0.5695050580200022)--cycle, qqqqcc); draw(arc((-0.7454801878199087,0.5695050580200022),0.2293757649057543,142.62220134846024,214.116625638157)--(-0.7454801878199087,0.5695050580200022)--cycle, qqqqcc); /* draw figures */ draw(circle((0.,0.), 2.)); draw((-3.3882509084590824,2.5884336857446697)--(0.8627441091058572,-1.8043482485936968)); draw((0.8627441091058572,-1.8043482485936968)--(-2.1577330600119065,-1.2791291579274162)); draw((-2.1577330600119065,-1.2791291579274162)--(-1.775081340838163,0.9214587529608635)); draw((-1.421188513146193,-1.4072040399676633)--(-3.3882509084590824,2.5884336857446697)); draw(circle((-1.7015644441457536,-0.8376844101109996), 0.6347938810952272)); draw((-0.7454801878199087,0.5695050580200022)--(-2.1577330600119065,-1.2791291579274162)); draw((-3.3882509084590824,2.5884336857446697)--(0.,0.)); draw((-0.7454801878199087,0.5695050580200022)--(-1.421188513146193,-1.4072040399676633)); draw((-0.7454801878199087,0.5695050580200022)--(-1.9819403751453128,-0.2681647802543354)); draw((-0.7454801878199087,0.5695050580200022)--(-0.262396510546413,1.982712301685009), linetype("2 2")); draw((0.,0.)--(-0.262396510546413,1.982712301685009)); draw((0.,0.)--(-1.9819403751453128,-0.2681647802543354)); draw((-0.262396510546413,1.982712301685009)--(-1.9819403751453128,-0.2681647802543354)); draw((-1.9819403751453128,-0.2681647802543354)--(-1.3955963959318056,-0.28149485164366533)); draw((-1.3955963959318056,-0.28149485164366533)--(-1.421188513146193,-1.4072040399676633)); /* dots and labels */ dot((0.,0.),linewidth(3.pt) + dotstyle); label("$O$", (0.051059991472206086,0.07317522097435145), NE * labelscalefactor); dot((0.8627441091058572,-1.8043482485936968),linewidth(3.pt) + dotstyle); label("$B$", (0.899750321623497,-2.071488180894451), NE * labelscalefactor); dot((-1.421188513146193,-1.4072040399676633),linewidth(3.pt) + dotstyle); label("$C$", (-1.53,-1.72), NE * labelscalefactor); dot((-1.775081340838163,0.9214587529608635),linewidth(3.pt) + dotstyle); label("$A$", (-2.2,0.8), NE * labelscalefactor); dot((-1.9819403751453128,-0.2681647802543354),linewidth(3.pt) + dotstyle); label("$D$", (-2.33,-0.31676357936543076), NE * labelscalefactor); dot((-3.3882509084590824,2.5884336857446697),linewidth(3.pt) + dotstyle); label("$F$", (-3.343701329132957,2.653652576164087), NE * labelscalefactor); dot((-2.1577330600119065,-1.2791291579274162),linewidth(3.pt) + dotstyle); label("$E$", (-2.33,-1.58), NE * labelscalefactor); dot((-0.7454801878199087,0.5695050580200022),linewidth(3.pt) + dotstyle); label("$Q$", (-0.625598514999769,0.6007394802575863), NE * labelscalefactor); dot((-1.3955963959318056,-0.28149485164366533),linewidth(3.pt) + dotstyle); label("$P$", (-1.42,-0.62), NE * labelscalefactor); dot((-0.262396510546413,1.982712301685009),linewidth(3.pt) + dotstyle); label("$R$", (-0.4076915383393025,2.091681952144989), NE * labelscalefactor); dot((-1.748171009353397,-0.7430138238330719),linewidth(3.pt) + dotstyle); label("$Z$", (-1.82,-0.6), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] We will use the following lemma: Lemma:The quadrilateral $OQDC$ is cyclic. Proof: Let $R$ be the second intersection of $CQ$ with the circle. We have $\angle{DQX}=\angle{CQY}\Rightarrow \angle{DQX}=\angle{RQX}\Rightarrow \angle{DQO}=\angle{RQO}$. The sine law gives $\frac{OQ}{\sin \hat{ODQ}}=\frac{OD}{\sin \hat{DQO}}$ and $\frac{OQ}{\sin \hat{QRO}}=\frac{OR}{\sin \hat{RQO}}$. Since $OD=OR$ and $\angle{DQO}=\angle{RQO}$,we finally obtain $\sin \hat{ODQ}=\sin \hat{ORQ}$. Both angles are acute,thus $\angle{ODQ}=\angle{ORQ}$. Hence $\triangle{ODQ}=\triangle{ORQ}\Rightarrow QD=QR\Rightarrow \angle{DRQ}=\angle{RDQ}\Rightarrow \angle{DQC}=2\angle{DRQ}=2\angle{DRC}=\angle{DOC}$, and we are done. $\rule{430pt}{1pt}$ From the lemma we obtain that the power of $F$ with respect to the circle $\odot{(A,B,C,D)}$ is equal to $FQ\cdot FO$, whence it follows that $Q$ lies on the polar of $F$ with respect to this circle.$E$ also lies on this line, hence $EQ$ is the polar of $F$ with respect to $\odot{(A,B,C,D)}$. Thus,$EQ\perp FO$,and,since $QF$ is the external bisector of $\angle{DQC}$,it follows that $QE$ is its internal bisector. Consequently,the pencil $Q.(F,E/D,C)$ is harmonic.Suppose that $Z\equiv QE\cap CD$. Then,the quadruple $(F,Z/D,C)$ is harmonic,which gives that the pencil $P.(F,Z/D,C)$ is harmonic.Since $PD$ is the internal bisector of $\angle{FPD}$,it follows that $PC$ is its external bisector. Thus $\angle{DPC}=90^{\circ}\Rightarrow \angle{DEC}=90^{\circ}$ and we are done. Note:The points $X,Y$ are not necessary for the proof.

Let $S$ be on $CD$ such that $(F,S;D,C)=-1$. $ES$ passes through $AC\cap BD$ and by Brocard we have $ES\perp FO$. Also $FO$ is the external angle bisector of $\measuredangle CQD$ hence $QS$ is its internal angle bisector which implies $QS\perp FO$. Combining these two yields $Q,E,S$ are collinear. $-1=(PE,PF;PC,PD)$ and $PD$ is the angle bisector of $\measuredangle EPF$ hence $\measuredangle CPD$. Thus, $\measuredangle AEB=\measuredangle DEC=90$ as desired.$\blacksquare$