(cases)AE = AF∠ EAM = ∠ FAMAM = AM (cases)\∴△AEM≌△AFM (SAS),∴EM = FM;∵四边形ABCD是正方形,∴BC = CD = 4, ∠BCD = 90°,设DM = x, 则MC = CD - DM = 4 - x,CE = BC - BE = 4 - 1 = 3,EM = FM = FD + DM = 1 + x,在Rt△MCE中, 根据勾股定理,...
∴∠EAM = ∠FAM , 又∵AM = AM ∴△AEM ≌△AFM(SAS) ∴EM = FM, 设DM = x,EM = FM = DF + DM = x + 1, CM = CD-DM = 4-х, 在Rt△CEM中,由勾股定理得EM^2=CE^2+CM^2; 即(x+1)^2=3^2+(4-x)^2 解得x=(12)/5∴ DM =(12)/5故答案为:D。反馈...
∵AM平分∠EAF, ∴∠EAM=∠FAM 又∵AM=AM . ∴△AEM≅△AFM(S.A.S) .). ∴EM=FM 设 DM =x.则 MC =CD -DM =4-x,CE =BC -BE = 4-1=3.E M =FM =FD +D M =1+x.在 Rt△MCE中 根据勾股定理,得 EM=MC^2+CE^2 ,即(1+x)=(4 x)"+3",解得 x-(12)/5 ...