Let $M,N$ be the midpoints of $AB,AC$. As $E$ is the intersection between $AD$ and the perpendicular bisector of $AB$, we get that $B,D,E,M$ are concyclic. Similarly $C,D,F,N$ are concyclic. Then $T=\odot BDEM\cap \odot CDFN\neq D$. By Miquel on $ABC$, we have that $A,M,N,T$ are concyclic. Then $MN$ is tangent to $\odot BDEM$ and $\odot CDFN$ because $\angle EMN=\angle BMN-\frac{\pi}{2}=\frac{\pi}{2}-\beta=\angle EAB=\angle EBM$ and similarly $\angle FNM=\angle FCN$ Hence $\angle TCA=\angle TCN=\angle TNM=\angle TAM=\angle TAB$, $\angle TAC=\angle TAN=\angle TMN=\angle TBM=\angle TBA$ thus $\triangle TBA \sim \triangle TAC$ hence $|TA|^2=|TB|\cdot |TC|$