Because triangle ABE is similar to triangle AEF, AE⊥EF and ABCD are rectangles, and angle BAE= angle EAF.

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.