$ (1)$证明：连接$AC$
在$∆AEC$和$∆AF C$中
$ \begin {cases}{AE=AF}\\{CE=CF}\\{AC=AC}\end {cases}$
∴$∆AEC≌∆AF C(\mathrm {SSS})$
∴$∠EAC=∠F AC$
∵$CB⊥AB，$$CD⊥AD，$
∴$∠ABC=∠ADC=90°$
在$∆ABC$和$∆ADC$中
$ \begin {cases}{∠BAC=∠DAC}\\{∠ABC=∠ADC}\\{AC=AC}\end {cases}$
∴$∆ABC≌∆ADC(\mathrm {AAS})$
∴$CB=CD$
$(2)$解：∵$AE=AF=8，$$CD=CB=6$
由$(1)$知$∆AEC≌∆AF C$
∴$S_{四边形AECF}=S_{△AEC}+S_{△AF} C=2×S_{△AEC}$
$=2×(\frac {1}{2}×AE×CB)=2×(\frac {1}{2}×8×6)=48$
