$(1)∵矩形ABCD沿对角线AC折叠$
$∴DA=BC=EC，∠D=∠B=∠E=90°$
$在△DAF和△ECF中，$
$\left\{ \begin{array}{l}{∠DFA=∠EFC}\\ {∠D=∠E} \\ {DA=EC} \end{array} \right.$
$∴△DAF≌△ECF$
$(2)∵△DAF≌△ECF$
$∴∠DAF=∠ECF=40°.$
$∵四边形ABCD是矩形， $
$ ∴∠DAB=90°.$
$∴∠EAB=∠DAB-∠DAF=50°$
$由折叠，知∠EAC=∠CAB.$
$∴∠CAB=25°$
