∴DP/BQ=AP/AQ,PE/QC=AP/AQ
∴DP/BQ=PE/QC
(2) ① Obviously BG = DG = DE = FG = EF = CF
It can be obtained from (1): dm/bg = mn/fg = ne/cf.
∴dm=mn=ne= 1/3de= 1/9bc=√2/9
② From the above, it can be seen that DM/BG = Mn/FG, and Mn/FG = NE/CF.
∴DM/MN=BG/FG,NE/MN=CF/FG
∫∠BAC = 90
∴∠B+∠C=90
∠∠b+∠BDG = 90°
∴∠C=∠BDG
∴△BDG∽△EFC
∴BG/EF=DG/CF
∴BG/FG=FG/CF
∴DM/MN=NE/MN
∴MN？ =DM×EN