$證明:(1)∵四邊形ABCD為菱形$
$∴AB=AD,∠B=∠D$
$又∵AE⊥BC,AF⊥CD$
$∴∠AEB=∠AFD=90°.$
$在△AEB和△AFD中$
$\begin{cases}{ ∠AEB=∠AFD }\ \\ {∠B=∠D\ } \\{AB=AD } \end{cases}$
$∴△AEB≌△AFD(AAS),∴AE=AF$
$(2)∵四邊形ABCD為菱形，∴∠B+∠BAD=180°$
$∵∠B=60°， ∴∠BAD=120°$
$又∵∠AEB=90°,∠B=60°,∴∠BAE=30°$
$由(1)知△AEB≌△AFD,∴∠BAE=∠DAF=30°$
$∴∠EAF=120°-30°-30°=60°$
$∵AE=AF,∴△AEF是等邊三角形$
$\ ∴∠AEF=60°$
