Let the area of ABCD be 100 . Side of ABCD =10 Area of EFGH is 62.5 ⇒ Side of EFGH = √62.5 Triangles AEH, BFE, CGF and DHG are congruent by ASA. Let
AE=BF=CG=DH=x;EB=FC=DG=AH=10−xx
AE2+AH2=EH2 x2+(10−x)2=(√62.5)2 Solving, x=2.5 or 7.5 Since it's given that CG is longer than EB,CG=7.5 and EB=2.5. Therefore, EB:CG=1:3