[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */import graph; size(16cm); real labelscalefactor = 0.3; /* changes label-to-point distance */pen dps = linewidth(0.1) + fontsize(11); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -22.610054216611292, xmax = 14.340581442303446, ymin = -9.398882860686594, ymax = 14.389904736212625; /* image dimensions */pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */draw(circle((-3.74,0.43), 5.704804992285013), linewidth(2) + wrwrwr); draw((xmin, -3.044943820224719*xmin + 7.325505617977529)--(xmax, -3.044943820224719*xmax + 7.325505617977529), linewidth(2) + wrwrwr); /* line */draw((xmin, -3.044943820224719*xmin-21.879333599960077)--(xmax, -3.044943820224719*xmax-21.879333599960077), linewidth(2) + wrwrwr); /* line */draw((-3.74,0.43)--(-5.549940947611276,-4.980075208919003), linewidth(2) + wrwrwr); draw((xmin, 2.9890893490525827*xmin + 11.60919416545666)--(xmax, 2.9890893490525827*xmax + 11.60919416545666), linewidth(2) + wrwrwr); /* line */draw((xmin, 0.7502889467283119*xmin + 10.019822119515986)--(xmax, 0.7502889467283119*xmax + 10.019822119515986), linewidth(2) + wrwrwr); /* line */draw((xmin, 0.3284132841328413*xmin + 1.6582656826568265)--(xmax, 0.3284132841328413*xmax + 1.6582656826568265), linewidth(2) + wrwrwr); /* line */draw(circle((-12.636832312768592,0.3986401969404606), 8.896887581551832), linewidth(2) + wrwrwr); draw((xmin, -283.70179161766924*xmin-1060.6147006500842)--(xmax, -283.70179161766924*xmax-1060.6147006500842), linewidth(2) + wrwrwr); /* line */draw((xmin, -283.70179161766924*xmin-1579.508265418503)--(xmax, -283.70179161766924*xmax-1579.508265418503), linewidth(2) + wrwrwr); /* line */draw((xmin, -3.044943820224719*xmin-21.879333599960077)--(xmax, -3.044943820224719*xmax-21.879333599960077), linewidth(2) + wrwrwr); /* line */ /* dots and labels */dot((-3.74,0.43),dotstyle); label("$A$", (-3.646473264230338,0.6925439388704359), NE * labelscalefactor); dot((1.68,2.21),dotstyle); label("$B$", (1.7753987180509476,2.4565979809523846), NE * labelscalefactor); dot((-8.405059103946876,3.7135991772264756),dotstyle); label("$C$", (-8.290086110298999,3.9612323109634584), NE * labelscalefactor); dot((-5.549940947611276,-4.980075208919003),linewidth(4pt) + dotstyle); label("$D$", (-5.436469277519375,-4.781211985825022), NE * labelscalefactor); dot((-0.7099212794006214,9.487176030534478),linewidth(4pt) + dotstyle); label("$E$", (-0.6112626330011015,9.694407947729792), NE * labelscalefactor); dot((-19.81995449895605,-4.850870665708813),linewidth(4pt) + dotstyle); label("$F$", (-19.704553441417495,-4.651502129789585), NE * labelscalefactor); dot((-5.588034667857405,5.827181474287961),linewidth(4pt) + dotstyle); label("$P$", (-5.48835321993355,6.036590007530457), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let the said line be $q$, and let the point where $q$ intersects the circumcircle of $AFD$ be P. First, we prove that CF intersects q at $P$. $PFA=DFA$, as A is the midpoint of arc $DP$. Also, since CD is parallel to line $p$, which is perpendicular to $AF$. Since CD is chord of circle $\mathcal{K}$, C is a reflection of D across $AF$, implying $DFA=CFA$, which means $C$ lies on line $FP$ Now that we know this, we have $\angle CPD=\angle FAD=\frac{1}{2}\angle CAD=\angle CBD$, implying $P, B, C, D$ are concyclic, so $P$ lies on the circle $\mathcal{K}$

Let $G\equiv EC \cap \cal ( K)$ we ll prove that $DG$ is parallel to the tangent at $A$ ;as showed in the diagram $B$ is on the bisector of $CD$ so $\angle DAB =\angle DKC$ where $K\equiv AD\cap \cal (K)$ thus $AF\parallel KC$ hence by Reim's we deduce $GDAF$ is cyclic besides $\angle DFA =\angle AFC$ since $F$ is on the bisector of $CD$ so$\angle LAD = \angle DFA =\angle AFC =\angle GDA$ where $L$ a point on the tangent as indicated in the diagram. therefore $DG$ is parallel to the tangent at $A$ RH HAS

Attachments:

Let CE meet $\mathcal{K}$ at S. Claim1 : AFSD is cyclic. Proof : ∠DAF = 2∠DCB = 180 - ∠DBC = 180 - ∠DSF. Claim2 : DS and tangent at A to AFD are parallel. Proof : Let tangent at A meet CF at P. ∠SAP = ∠SDP = ∠DSA. we're Done.