Solution:
Let $\alpha=\frac{1}{2}\angle A$, $\beta=\frac{1}{2}\angle B$, and $\gamma=\frac{1}{2}\angle C$.
Assume $BEAR$ is cyclic.
Note that $\angle LIB=90^{\circ}-\alpha=\frac{1}{2}(180-\angle A)=\frac{1}{2}\angle ERB=\angle LRB$ so $LBRI$ is cyclic. Let its circumcircle be $\omega$. Consider $\angle BRI=90^{\circ}+\gamma=\angle BIA$ so $AI$ is tangent to $\omega$ at $I$. It follows that $\angle LIA=\angle LRI=90^{\circ}-RLI=90^{\circ}-IBR=90^{\circ}-\beta$. But we also have $\angle LEA=\frac{1}{2}\angle REA=\frac{1}{2}(180^{\circ}-\angle B)=90^{\circ}-\beta$ so $LAEI$ is concyclic. But clearly, $AFIE$ is cyclic, so we must have $ALFIE$ cyclic. Now it follows that $LF$ and $LE$ are reflections of each other wrt. $CI$ but $LT$ and $LE$ are also reflections of each other wrt. $CI$, hence $L$, $F$, and $T$ colinear.
Assume $L$, $F$, and $T$ colinear. Then $\angle LFA=\angle BFT=\angle BTF=\angle LEA$ where the last equality follows from symmetry. But note that $\angle REA=\frac{1}{2}\angle LEA=\frac{1}{2}\angle BTF=\frac{1}{2}(90^{\circ}-\beta)=180^{\circ}-\angle B$ so $BEAR$ is cyclic.