(1) Let PM be perpendicular to BC and M, M, and PN be perpendicular to CA and N..

It is easy to prove that △PME and △PNF are congruent, so PF=PE.

CP=√2, so CN=CM=PN=PM= 1.

FN= 1-x, CG is parallel to PN, so PN/CG = NF/CF.

CG=PN*CF/NF=x/( 1-x)

CE = CM+ME = CM+NF = 1+ 1-x = 2-x

EG=CG+CE

y=x/( 1-x)+2-x=(x^2-2x+2)/( 1-x)= 1-x+ 1/( 1-x)

Domain 0