Let the tangent from $E$ to $ACD$ touch at $M$. Let the tangent from $E$ to $BCD$ touch at $T$ . If we prove $\angle TEB = \angle BAC$ we're done. Claim: $MECT$ cyclic. Proof: Redefine $T$ as the intersection of the circle $MEC$ with $BCD$. By Reim $M,T,D$ collinear. $\angle ETC=\angle EMC = \angle MAC = \angle MDC = \angle TDC = \angle TBC$, so $ET$ is tangent to $BCD$ since $\angle ETC=\angle TBC$. Some more angles $\angle TEC = \angle TMC = \angle DMC = \angle DAC = \angle BAC$ done

Did asymptote for fun, rip wasted time Let $R$ be the tangency point to $(ACD)$ from $E$ and let $T=DR\cap (BDC)$. Let $P=AB\cap TE$. Firstly, we get $$\measuredangle CER=\measuredangle BER=\measuredangle CBD=\measuredangle CTD=\measuredangle CTR,$$thus $CETR$ is cyclic. Now, $\measuredangle ETC=\measuredangle ERC=\measuredangle RDC=\measuredangle TDC$, which means that $TE$ is in fact tangent to the circumcircle of $\triangle BCD$. By Miquel's theorem on $\triangle PDT$, we conclude that $ACEP$ is cyclic. Hence, $\measuredangle PEB=\measuredangle PEC=\measuredangle PAC=\measuredangle BAC$ and $\measuredangle CBA=\measuredangle EBP$, thus we have $\triangle ABC\sim\triangle EBP$. [asy][asy] size(10cm); import olympiad; import geometry; pair A=dir(95); pair D=dir(160); pair C=dir(0); dot("$A$", A, dir(A)); dot("$C$", C, dir(45)); dot("$D$", D, dir(D)); pair M=(A+D)/2; pair B=-1.5A+2.5D; dot("$B$", B, dir(B)); draw(A--B--C--cycle); draw(circumcircle(A,D,C)); pair O=(0,0); pair X=intersectionpoints(circle(O,1),4O-3M--O)[0]; dot("$R$", X, dir(250)); pair E=intersectionpoint(perpendicular(X,line(X,O)),line(B,C)); dot("$E$", E, dir(E)); draw(C--E); pair N=circumcenter(B,D,C); path c= circle(N,circumradius(B,D,C)); draw(c); pair Z=(N+E)/2; path k=circle(Z,abs(Z-E)); pair T=intersectionpoints(c,k)[0]; dot("$T$", T, dir(T)); pair P=extension(B,D,T,E); dot("$P$", P, dir(P)); draw(A--P); draw(T--P); draw(X--E); draw(D--T); [/asy][/asy]

Let the tangent from $E$ to $(ACD)$ meet $(ACD)$ at $F$. Let $DF\cap (BCD) = G$. We have $\angle CEF = \angle BEF = 180 - \angle EBA = 180 - \angle DBC = \angle CGD \implies EFGC$ is cyclic. Also $\angle EGC = \angle EFC = \angle FAC = \angle FDC = \angle GDC \implies EG$ is tangent to $(BCD)$. Finally $\angle BEG = \angle CEG = \angle CFG = \angle CFD = \angle CAD$ and we are done.$\square$

Let the line through $E$ parallel to $AB$ meets $(ACD)$ at $F$ . now , looking the diagram from $\triangle FAD$ , the main problem is equivalent to this: "In an isosceles triangle $\triangle ABC$ with $AB = AC$ , an arbitrary line intersects with $BC , (ABC) , AA$ at $X , Y ,Z$ respectively (suppose that $A,Y$ are in one side of $BC$ and WLOG $Y$ is on smaller arc $AB$) . if the tangent from $Z$ to $(XYC)$ meets $BC$ at $W$ prove that $BYZW$ is cyclic"

and we are done.... note:$AA$ means tangent from $A$ to a circle

Assume ES to be the tangent to circumcircle of ADC and EK to be the tangent to the circumcircle of BCD. Since ES is parallel to AB we have the equation for these angles: DKC=180-DBC=SEC. which results in SKCE being cyclic. then EKC=ESC. and from the tangents we have: ESC=SDC , EKC=KDC so KDC=SDC so S,K,D are on the same line. It is easy to show that triangles ABC and SCK are similar. now if EK intersects AB at F we just have to prove that triangles SCK and BEF are similar which using cyclic quadrilaterals is easy.

Let $EX,EY$ be tangents to $ADC,BDC$. Let $DY$ meet $ADC$ at $X'$. Let $EY,AB$ meet at $F$. Claim $1: X'$ is $X$. Proof $:$ Note that $\angle X'EC = 180 - \angle B = \angle DYC$ so $YX'EC$ is cyclic so $\angle CX'E = \angle CYE = \angle CDY = \angle CDX'$ which implies that $CX'$ is tangent to $ADC$ so $X'$ is $X$. Now we have $\angle DAC = \angle DXC = \angle YXC = \angle YEC = \angle FEC$ so $AFEC$ is cyclic so $ABC$ and $EBF$ are similar as wanted.