$(1)$证明：∵$AC$平分$∠BAD$
∴$∠BAC = ∠DAC$
$ $又$CB\perp AB，$$CD\perp AD$
∴$∠B = ∠D = 90°$
在$\triangle ABC$和$\triangle ADC$中
$\begin {cases}∠B=∠D\\∠BAC=∠DAC\\AC=AC\end {cases}$
∴$\triangle ABC≌\triangle ADC(\mathrm {AAS})$
$ (2)$解：由$(1)$得$\triangle ABC≌\triangle ADC$
∴$CB = CD，$$S_{\triangle ABC} = S_{\triangle ADC}$
又$CD = 3，$∴$CB = 3$
又$AB = 4，$$CB\perp AB$
∴$S_{\triangle ABC}=\frac 12\ \mathrm {A}B·CB = 6$
∴$S_{四边形ABCD}=S_{\triangle ABC}+S_{\triangle ADC}$
$=2S_{\triangle ABC}=12$
