Question

求解答 第6题

Answer 1

1.ABC为等腰直角，<ABC=45，<MBD =<DBC = 22.5
BD = BC cos22.5
BM=BD cos22.5 = BC(cos22.5)^2 = BC (1+cos45)/2
AB = BC cos 45
BM/(AB+BC) = BC(1+cos45)/(2BC(1+cos45)) = 1/2
2.AM =BM - AB = BC(1-cos45)/2, BC-AB = BC (1 - cos45)
AM/(BC-AB) = 1/2

Answer 2

解:延长CD交BM的延长线于F.∠FBD=∠CBD,BD=BD,∠BDF=∠BDC=90°,则⊿BDF≌⊿BDC,BF=BC;DF=DC.DM与CA都垂直于BF,则:DM∥CA,FM/MA=DF/DC=1,FM=MA.∴BM/(AB+BC)=BM/[(BM-AM)+(BM+FM)]=BM/(2BM)=1/2;AM/(BC-AB)=AM/(BF-AB)=AM/(2AM)=1/2.