If e is the vertical line of AF and the vertical foot is h, then BE=EH and AB=AH=3.
According to the topic, triangle EHF≌ triangle ECF, EC=EH.
So BE=EC=EH=2.
Let DF=x, then FC=HF=3-x, af = ah+HF = 6-X.
In right triangle ADF, AF? =DF? +AD? ,(6-x)? =x? + 16,x=5/3。
AF= 13/5
The area of triangle AEF =AF*EH/2= 13/5.