Question

分别以三角形ABC边AB,BC,AC为斜边作等腰直角三角形DAB,EBC,FAC.证：AE=DF、AE垂直DF.

Answer 1

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