Lemma: $F$ is the midpoint of $BC$. From $\angle CDF = \angle BFE$, we find $\triangle FDC \sim \triangle EFB$. So $BF \cdot FC = BE \cdot CD$. By Power of a Point, $BT^2 = BE \cdot BA$. This becomes $BF^2 = BE \cdot CD$. So $BF = FC$. Note that $\angle DFC$ and $\angle BFE$ are complementary, so $\angle DFE = 90^{\circ}$. Now, we compute the ratio of the legs of $\triangle DEF$ and $\triangle DFC$ to each other. Let $BE = a$ and $BF = R$. Note that $EF = \sqrt{a^2 + R^2}$ and $DF = \sqrt{R^2 + \tfrac{R^4}{a^2}}$, or $\tfrac{R}{a}\sqrt{a^2 + R^2}$. Thus, in $\triangle DEF$ the ratio is $\tfrac{a}{R}$. In $\triangle DFC$, it is $\tfrac{R}{\frac{R^2}{a}}$, or $\tfrac{a}{R}$. They are the same, so by ASA Similarity $\triangle DEF \sim \triangle DFC$. Thus $\angle EDF = \angle CDF$. $\blacksquare$