$證明:(1)∵CF⊥AD,BE⊥AD,∴∠CFD=∠BED=90°\ $
$∵AD是△ABC的中線，∴BD=CD$
$在△CFD和△BED中$
$\begin{cases}{ ∠CFD=∠BED }\ \\ { ∠CDF=∠BDE } \\{CD=BD } \end{cases}\ $
$∴△CFD≌△BED(AAS),∴CF=BE$
$(2)解:∵S_{△ACF}=28,S_{△CFD}=12, ∴S_{△ACD}=S_{△ACF}+S_{△CFD}=40$
$∵BD=CD,∴S_{△ABD}=S_{△ACD}=40$
$由(1)得:△CFD≌△BED,∴S_{△CFD}=S_{△BED}=12$
$∴S_{△ABE}=S_{△ABD}+S_{△BED}=40+12=52$
