解：設(shè)正方形的邊長(zhǎng)為?$2a，$?易得?$ED=a，$??$CE=\sqrt{5}a$?
過(guò)點(diǎn)?$H$?作?$MN//AD，$?分別交?$AB，$??$CD$?于點(diǎn)?$M，$??$N$?
則?$△CNH∽△CDE$?
∴?$\frac {NH}{ED}=\frac {CH}{CE}$?
∵?$CH= 2a，$??$ED=a，$??$CE=\sqrt{5}a$?
∴?$NH=\frac {2\sqrt{5}}{5}a$?
∴?$MH= 2a-\frac {2\sqrt{5}}{5}a$?
由上述可得，?$∠HMG =∠CNH=∠CHG= 90°$?
得?$∠MHG=∠NCH，$??$△HMG∽△CNH$?
∴?$△HMG∽△CDE$?
∴?$\frac {HG}{EC}=\frac {MH}{CD}$?
則?$HG=\frac {\sqrt{5}}{2}(2a-\frac {2\sqrt{5}}{5}a)=(\sqrt{5}-1)a$?
∴?$BG= HG=\frac {\sqrt{5}-1}{2}AB$?
∴點(diǎn)?$G $?為?$AB$?的黃金分割點(diǎn)
