In Rt△A'EB, let A'E be x and EB=√3×x, then the length of the diamond is (√ 3+1) x.

A'B=2x，BD ' =(√3+ 1)x-2x =(√3- 1)x = D ' h，

In Rt△HCF, let FC=t, FH=√3×t, FC+FH=(√3+ 1)t=2x.

T=(√3- 1)x=FC，FH = (3-√ 3) x .

FC:FD = FC:FD ' = FC:(FH+HD ')=(√3- 1)x:[(3-√3)x+(√3- 1)x]=(√3- 1)/2

Note that the known conditions in the problem are marked in the figure, and Rt△A'EB and Rt△HCF are obtained. According to the quantitative relationship, the results are deduced step by step.