See that $\Delta A'BC'\sim\Delta A'B'C$ and, if $F'$ and $E'$ are the projections of $A'$ onto $AB, AC$, $E', F'$ verify the given relation and $AF'A'E'$ is cyclic, since $\angle AF'A'=\angle A'E'A=90^\circ$. Next, see that for any $F', E'$ satisfying the relation, we get $\triangle A'F'F\sim\triangle A'E'E$, i.e. $\angle FA'F'=\angle EAE'$ and subsequently $\angle FA'E=180^\circ-\hat A$, done. Best regards, sunken rock