详细一点：原贴CD应为DF

连接FE,把ABCD分成四块，
area of the squre: AB^2
Area of ABE:  1/2xABxBE
Area of ADF: 1/2xADxDF=1/2xABxDF
Area of CFE: 1/2xCExCF=1/2x(AB-BE)x(AB-DF)
AEF has same height of ADF if let AE as the base so
Area of AEF: 1/2xAEXDF

根据面积相加(x2)的关系得到

2AB^2-ABxBE-ABxDF-(AB-BE)x(AB-DF)=AExDF

化简=> (AE+BE)xDF=AB^2

and AB^2=AE^2-BE^2=(AE+BE)x(AE-BE)

带进去刚好剩下：AE = BE+DF