First, we know that since $\angle AEH=\angle ADH = 90^{\circ}$, $AEHD$ is cyclic. Thus $\angle DAH = \angle DEH$. We also know that be alternate interior angles and the given information that $B,E,D,G$ are collinear , $\angle DEH = \angle GEF = \angle AGE = \angle AGB$. Note that since $\angle AGT = \angle ABT = 90^{\circ}$, $ABTG$ is cyclic and $\angle AGB = \angle ATB$. Thus $\angle DAH = \angle ATB$. Note that $\angle ATB = \angle DAT$, thus $\angle DAH = \angle DAT$ and we are done. $\blacksquare$

It is easy to see that $AEHD$ and $ABTG$ are both cyclic, See that, $\angle ATG=\angle ABG=\angle ABD=\angle BDH=\angle EAH$ , and since $AE\parallel GT$ it follows that $A,H,T$ are collinear.

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