Problem 63

We will prove every result except for $\fbox{B}$.
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 $\fbox{B}$.