2 1, solution: see the figure below (1)AB=AE, BE = AB-AE = 0;

(2)ae=ad/cosa=[ac/cos(π/4-a)]/cosa=ac/cosacos(π/4-a)=4√2/[cos^2a+( 1/2)sin2a)；

ae'=-4√2[2cosa(-sina)+cos2a]/[(cos^2a+( 1/2)sin(2a)]^2

=4√2[sin(2a)-cos(2a)]/[cos^2a+sin(2a)]^2=0； Namely:

sin(2a)-cos(2a)= √{[ 1-cos(4a)]/2 }-√{[ 1+cos(4a)/2 }

= { √{[ 1-cos(4a)]/2 }-√{[ 1-cos(4a)/2 } } { √{[ 1-cos(4a)]/2 }+√{[ 1-cos(4a)/2 } }/{ √{[ 1-cos(4a)]/2 }+√{[ 1+cos(4a)/2 } }

=(√2/2){[ 1-cos(4a)]-[ 1+cos(4a)}/{ √[ 1-cos(4a)]+√[ 1+cos(4a)]]

=-√2 cos(4a)/{ √[ 1-cos(4a)]+√[ 1+cos(4a)]} = 0； 4a=π/2。 ? a=π/8

aemin=4√2/[cos^2(π/8)+( 1/2)sin(π/4)]=4√2/{[ 1+cos(π/4)]/2+( 1/2)sin(π/4)]}

=4√2/[( 1+√2/2)/2+( 1/2)(√2/2)]=8√2/[( 1+√2]=8(2-√2)。