DF⊥AB,EG⊥AC

AB=AC

∴∠ABC=∠ACB

△DBA?△AEC

DF=AF=BF=AG=CG=EG

∫M is the midpoint of BC, that is, BM=CM.

BF=CG，∠FBM=∠GCM

∴△BMF≌△CMG(SAS)

∴FM=MG

F, m and g are the midpoint.

∴ median theorem: FM= 1/2AC=AG=EG, MG= 1/2AB=AF=DE, fm∨AC.

∴MG=EG,∠GME=∠GEM=y

∠FMG+y+∠CME+X= 180

∵ easily obtained: △ AFG △ MFG, ∠ FMG = ∠ A = 180-2x.

∴∠CME=X-y

∠∠FGC+y =∠CME+X

That is 90+y-x = ∠ CME.

∴90 -(X-Y)=∠CME

90 -∠CME=∠CME

∠CME=45

∫∠CAE = 45

∴∠EMC=∠CAE