Are you sure that the problem is correct $?$

Oopss, I'm very sorry I have now corrected the angles in the summation.

Let $\angle{ACE}=x\angle{GDC}=y,\angle{EAF}=z,\angle{CAF}=t$, we have $t=180-2x-y-z$ and let $\angle{ABG}=l$ Since $\angle{AGB}=\angle{ACG}=x$, let $T+AC\cap BG$, we get that $AT\cdot AC=AG^2$ So $(\frac{AG}{AB})^2=\frac{AT}{AB}\cdot \frac{AC}{AB}$ give us $\frac{sin^2(l)}{sin^2(x)}=\frac{sin(l)sin(2x+2y)}{sin(l+x+y)sin(x+y)}$ So $sin(l)sin(l+x+y)=2sin^2(x)cos(x+y)$ So $cos(x+y)-cos(x+y+2l)=2(sin(2x+y)-sin(y))sin(x)=(cos(x+y)-cos(3x+y))-(cos(x-y)-cos(x+y))$ So $cos(x+y)+cos(x+y+2l)=cos(x-y)+cos(3x+y)$ give us $2cos(x+y+l)cos(l)=2cos(2x)cos(x+y)$ And we have $1=\frac{AG}{GC}\cdot \frac{CG}{DG}\cdot \frac{DG}{AG} =\frac{sin(t)sin(x+2y+z+t)sin(z)}{sin(z)sin(y)sin(x)}$ So $sin(t)sin(x+y+2z+t)=sin(x)sin(y)$ give us $cos(x+y+2z)-cos(x+y+2z+2t)=cos(x-y)-cos(x+y)$ We get $cos(x-y)+cos(x+y+2z+2t)=cos(x+y)+cos(x+y+2z)$ give $2cos(x+y+z)cos(z)=2cos(x+z+t)cos(y+z+t)$ Put $t=180-2x-y-z$ give $2cos(x+y+z)cos(z)=2cos(x+y)cos(2x)$ So $cos(x+y+z)cos(z)=cos(x+y+l)cos(l)$ give $cos(x+y+2z)+cos(x+y)=cos(x+y+2l)+cos(x+y)$ and so $z=l$ So $AD$ tangent to $(AGB)$ as we want

$B$ is on the polar of the intersection $AC\cap EF$ thus $B,G,D$ are collinear.let $H$ the intersection of $\omega$ an the parallel of $GD$ through $F$ then $\widehat{HFC}=\widehat{HDC}$ and so $\widehat{HFA}=\widehat{ACE}$. since $BA$ tangent to $\omega$ we have $\widehat{EAB}=\pi-\widehat{ACE}=\pi -\widehat{HFA}=\pi -\widehat{BGA}$ which means that $EA$ is tangent to the circumcircle of ABG. R HAS

Pascal on $AAFCCE$ gives that $B,G,D$ are collinear. \[\angle BAC+\angle FCE=\angle GDC+\angle ECA+\angle FAD=\angle BGC+\angle GCA=180-\angle CAG-\angle AGB=\angle BAC+\angle GBA\]Thus $\angle GAD=\angle DBA$ as desired.$\blacksquare$