Let NH be perpendicular to CF in h, then NH=CH.

AM is perpendicular to MN, then: ∠NMH=∠BAM (both are complementary angles of ∠BMA);

And ∠B=∠MHN. If ⊿ mhn ∽ ABM, MH/HN = AB/BM = 2, that is, MH=2HN=2CH, then HN=CH.

(2)⊿mhn∽⊿abm； And NH=BM. So ⊿MHN≌⊿ABM (congruence of two triangles with similarity ratio 1), MH=AB.

If NH is parallel to EF, then CN/CE = NH/EF = NH/AB = NH/MH =1/2, so CN=( 1/2)CE, that is, N is the midpoint of CE.

2. Solution: When point M is on the side of BC, CM=m, then BM = BC-cm = 1-m Let NH be perpendicular to CF in H, then HN=CH.

(1) Take BG=BM on BA and connect GM, then AG = CM∠BGM=∠BMG=45, AGM = ∠ MCN =135;

And ∠GAM=∠CMN (authentication). So ⊿MHN≌⊿ABM, then ch = HN = BM =1-m.

If HN is parallel to EF, then CN/NE = ch/HF = (1-m)/(cf-ch) = (1-m)/m. 。

(2) When point M is on the BC extension line, the value of CN/NE in (1) will change.

The same similar method can prove ⊿MHN≌⊿ABM.

CM=m, then BM =1+m = HN; MH=AB= 1。

CN/NE = CH/FH =(CM+MH)/(MH-MF)=(m+ 1)/[ 1-( 1-m)]=( 1+m)/m。