1965 AMC 12 Problem 30

Use coordinates $C=(0,0), A=(a,0), B=(0,b)$

The tangent meets $CA$ at its midpoint; test the angle-bisector claim with the angle-bisector theorem

Put $C=(0,0),\ A=(a,0),$ and $B=(0,b).$ The second intersection of $AB$ with the circle of diameter $BC$ is $$D=\left(\frac{ab^2}{a^2+b^2}, \frac{a^2b}{a^2+b^2}\right).$$ The tangent at $D$ meets $CA$ at $F=(\frac{a}{2},0).$ Hence $DF$ bisects $CA;$ also $FC=FD=FA$ by equal tangents from $F,$ which supplies choices C and E, and the circle/right-triangle angles supply choice D.

If $DF$ bisected $\angle CDA,$ the angle-bisector theorem in triangle $CDA$ would require $\frac{CF}{FA}=\frac{CD}{DA}.$ The left side is $1,$ while $$\frac{CD}{DA}=\frac{ab}{a^2}=\frac ba,$$ which need not be $1.$ Thus choice B is not provable.

Therefore, the correct answer is B.