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Let $x=\angle A/2$. By the first angle condition, we obtain $\angle B=90^\circ+x,\angle C=90^\circ-3x$.

Since $AB=AF$, hence by the second angle condition, we obtain 

$\angle AFB=\angle ABF=90^\circ-(3x/2),\angle BAF=3x, \angle CDF=\angle BDA=90^\circ -x/2, \angle FAD=\angle FAB-\angle DAB=3x-2x=x$.

Let $D'$ be the reflection of $D$ over the midpoint of $AC$.

Since $AD'=AF$, hence $\angle AD'F=90-x/2=\angle CDF$.

$\therefore FD=FD'$ or $F$ lies on the perpendicular bisector of $DD'$ which is the perpendicular bisector of $AC$.

So, $\angle ABC=90^\circ+x, \angle FAD=\angle FAC=\angle FCA=x, \angle CFA=180^\circ - 2x = 360^\circ - 2\angle B$.

Since $\angle B$ is obtuse, hence $F$ is the circumcenter of $\triangle ABC$.