$20 \cdot 37$ Given a right triangle $\triangle ABC$ with a right angle at $B$, a circle is drawn with the diameter $BC$ intersecting the hypotenuse $AB$ at $D$. A tangent line through $D$ intersects $CA$ at $F$. These conditions are insufficient to prove:
(A) $DF$ bisects $CA$.
(B) $DF$ bisects $\angle CDA$.
(C) $DF=FA$.
(D) $\angle A=\angle BCD$.
(E) $\angle CFD=2\angle A$.
[Solution] Connect $C D$. Since $\angle A=\angle B C D$ (both are the complements of $\angle A C D$), and $F C=F D$ (tangents from a point to a circle are equal).
Therefore, $\angle D C F=\angle C D F$,
Thus, $\angle B C D=\angle F D A$ (complements of equal angles).
That is, $\angle A=\angle B C D=\angle F D A$,
Hence, $D F=F A=C F$.
In the right triangle $\triangle A C D$, $D F$ is the median of $A C$.
Also, $\angle C F D=\angle F D A+\angle A=2 \angle A$.
Therefore, it has been proven that $(A),(C),(D),(E)$ are correct. Hence, the answer is $(B)$.