Question

在正方形ABCD的BC和CD上分别有点E、F,角EAF=45度,BD分别交AE、AF于M、N,AG垂直EF于G,且BE=

Answer 1

延长CB至Q,使BQ=DF=6,∵AD=AB,BQ=DF,〈FDA=〈QBA=90°,∴RT△ADF≌RT△ABQ,∴〈BAQ=〈FAD,AQ=AF,∵〈FAE=45°,〈DAB=90°,∴〈DAF+〈EAB=45°,∴〈QAE=〈EAB+〈BAQ=〈EAB+〈DAF=45°,∴〈EAQ=〈FAE=45°,∵AE=AE（公用边）,∴△FAE≌△QAE,∴EF=EQ=EB+BQ=4+6=10,设正方形边长AB=x,在△EFC中,根据勾股定理,(x-6)^2+(x-4)^2=10^2,x^2-10x-24=0,(x-12)(x+2)=0,∴x=12,(舍去负根-2）,AB=12,根据勾股定理,AE=√（AB^2+BE^2)=4√10,在△MBE中,BM=3√2.BE=4,〈MBE=45°.根据余弦定理,ME=√10,AM=AE-ME=3√10,∵〈EMB=〈NMA,（对顶角相等）,〈MBE=〈NAM=45°,∴△AMN∽△BME,∴BM/AM=ME/MN,3√2/（3√10）=√10/MN,∴MN=5√2.