smc

We will prove every result except for $\boxed{B}$. By Thales' Theorem, $\angle CDB=90^{\circ }$ and so $\angle CDA=90^{\circ }$. $FC$ and $FD$ are both tangents to the same circle, and hence equal. Let $\angle CFD=\alpha$. Then $\angle FDC=\frac{180^{\circ }-\alpha }{2}$, and so $\angle FDA=\frac{\alpha }{2}$. We also have $\angle AFD=180^{\circ }-\alpha$, which implies $\angle FAD=\frac{\alpha }{2}$. This means that $CF=DF=FA$, so $DF$ indeed bisects $CA$. We also know that $\angle BCD=90-\frac{180^{\circ }-\alpha }{2}=\frac{\alpha }{2}$, hence $\angle A=\angle BCD$. And $\angle CFD=2\angle A$ as $\alpha =\frac{\alpha }{2}\times 2$. Since all of the results except for $B$ are true, our answer is $\boxed{DF\text{ bisects }\angle CDA}$.