$(1)$证明：∵$AC=CD$
∴$\widehat{AC}=\widehat{CD}$
∴$∠CAD=∠ABC$
∵$AB$是直径
∴$∠ACB=90°$
∴$∠ABC+∠BAC=90°$
∵$AE=AF，$$∠ACB=90°$
∴$∠FAC=∠CAD，$$CE=FC$
∴$∠FAC=∠CAD=∠ABC$
∴$∠FAC+∠CAB=90°$
∴$AF⊥AB$
又∵$AB$是直径
∴$AF $是$⊙O$的切线
$ (2) $连接$BD$
∵$cos ∠ABC=cos∠CAD=\frac {AC}{AE}=\frac {4}{5}，$$AC=4$
∴$AE=5$
∴$AF=AE=5，$$CE=\sqrt{AE^2-AC^2}=\sqrt{5^2-4^2}=3$
∴$CE=CF=3$
∵$tan F = \frac {AC}{CF}= \frac {AB}{AF}$
∴$ \frac {4}{3} =\frac {AB}{5}$
∴$AB=\frac {20}{3}$
∵$cos∠ABC=\frac {4}{5}=\frac {AB}{BF}$
∴$BF=\frac {25}{3}$
∴$ BE=BF-EF=\frac {25}{3} -(3+3)=\frac {7}{3}$
∵$∠CAD=∠CBD，$$∠AEC=∠BED$
∴$△ACE∽△BDE$
∴$\frac {AE}{BE}=\frac {EC}{ED} $
∴$\frac {5}{\frac {7}{3}}= \frac {3}{ED} $
∴$ED=\frac {7}{5}$
∴$AD=ED+AE=\frac {7}{5} +5=\frac {32}{5}$
∴$cos ∠BAD= \frac {AD}{AB} =\frac {\frac {32}{5}}{\frac {20}{3}} =\frac {24}{25}$
