用向量解答
设BC=(a,b),CA=(c,d)
则BA=(a+c,b+d)
BE=(a/2+b/2,-a/2+b/2)
FA=(c/2-d/2,c/2+d/2)
AD=(-a/2-c/2-b/2-d/2,a/2+c/2-b/2-d/2)
AE=AB+BE=(-a-c+a/2+b/2,-b-d-a/2+b/2)
=(-c-a/2+b/2,-d-a/2-b/2)
FD=FA+AD=(c/2-d/2-a/2-c/2-b/2-d/2,c/2+d/2+a/2+c/2-b/2-d/2)
=(-d-a/2-b/2,c+a/2-b/2)
根据上述AE,FD可得|EA|=|FD|,EA*FD=0
所以EA等于且垂直于FD