Proof: First of all, ΔABF = ΔAEC by SAS. This is because, AE = AB, AF = AC, and
∠BAF = ∠BAC + ∠CAF = ∠CAB + ∠BAE = ∠CAE.
ΔABF has base AF and the altitude from B equal to AC. Its area therefore equals half that of square on the side AC. On the other hand, ΔAEC has AE and the altitude from C equal to AM, where M is the point of intersection of AB with the line CL parallel to AE. Thus the area of ΔAEC equals half that of the rectangle AELM. Which says that the area AC² of the square on side AC equals the area of the rectangle AELM.
Similarly, the area BC² of the square on side BC equals that of rectangle BMLD. Finally, the two rectangles AELM and BMLD make up the square on the hypotenuse AB.