I really think that the poster should post the actual diagram..this problem isnt as hard if the diagram is given...
Ok lets see..
side of Square=S...side of Equilatral=A..
suppose point B and C are X..(btw you will quickly see that B and C are exactly X away)..i.e BE=CE..
similary BC=A => A^2=x^2+x^2 or 2x^2
area of ADB=1/2 (S)*(s-x) area of BCE=1/2 (X)*(X)
Ok, so we S^2 - 2xS + x^2=2X^2
x^2=2s^2-2xs or 2(s^2-xs)
area of ADB =1/2* (s^2-xs)
area of BCE=1/2*x^2 we know x^2 interms of S..
ADB/BCE = (s^2-xs)/2(s^2-xs)=1
ratio of BCE to ADB=2
took about 2 mins