As an auxiliary line BF//ED, BG is perpendicular to AC, and it can be proved that triangle AFB is similar to triangle ABC from the angle relationship. So ab 2 = af * AC,

That is, ab 2 = af, we assume AB=x,

Then af = x 2,

AG=x/2，

FG=x/2-x^2，

BC = FC = 1-x 2。 So GC=FC-FG= 1-x/2,

CD = Ce = BC/2 = ( 1-x 2)/2。 According to Pythagorean theorem in triangular BCG, GC 2+BG 2 = BC 2, and X 3-3x+1= 0 is obtained by sorting.

S

△ABC+2

S

△cde=bg/2+bg*cd/2=(√3/4)x[ 1+(( 1-x^2)/2]=√3/8(-x^3+3x)=√3/8* 1=√3/8