$(1)证明：∵AC平分∠BAD$
$∴∠BAC=∠DAC$
$∵CB⊥AB，CD⊥AD$
$∴∠B=∠D=90°$
$在△ABC和△ADC中，$
${{\begin{cases} { {∠B=∠D}} \\{∠BAC=∠DAC} \\ {AC=AC} \end{cases}}}$
$∴△ABC≌△ADC(AAS)$
$(2)∵△ABC≌△ADC$
$∴ BC=CD=3，S_{△ABC}=S_{△ADC}.$
$∵ S_{△ABC}=\frac{1}{2}AB·BC=\frac{1}{2}×4×3=6，$
$∴ S_{△ADC}=6，$
$∴S_{四边形ABCD}=S_{△ABC}+S_{△ADC}=12，$
$∴四边形ABCD的面积是12$
