$(2)解：∵Rt△ABC≌Rt△ADE$
$∴AC=AE，AD=AB，∠CAB=∠EAD$
$∴∠CAB-∠DAB=∠EAD-∠DAB$
$即∠CAD=∠EAB$
$在△CAD和△EAB中$
${{\begin{cases} {{AC=AE}}\\ {∠CAD=∠EAB}\\ {AD=AB} \end{cases}}}$
$∴△CAD≌△EAB(SAS)$
$∴CD=EB，∠ADC=∠ABE$
$∵∠ADE=∠ABC$
$∴∠CDF=∠EBF $
$∵∠DFC=∠BFE$
$在△CDF和△EBF中$
${{\begin{cases} {∠DFC=∠BFE } \\ {∠CDF=∠EBF } \\ {CD=EB} \end{cases}}}$
$∴△CDF≌△EBF(AAS)$
$∴CF=EF$
