This is an Olympic math problem in primary school, a typical rotation problem in the collection.

Please rotate △ADF 90 counterclockwise around point F, so that DF and EF coincide, and AD and BE are in a straight line. △ADF and △BEF form a new △ABF, where ∠ AFB = 90.

So s △ ABF =12×18×1/2 =108.

That is to say, the areas of △ADF and △FEB add up.

Of course, △BEF can also rotate 90 clockwise around point F.