Question

求解第二题

Answer 1

三角形ABC是以BC为斜边的等腰直角三角形，所以，BC=AB*√2=AC*√2
∠AGB=∠EGC
RT△AGB∽RT△FGC
FG/AG=CG/BG
FG=AG*CG/BG...................................................................(1)
AE平分∠BAC，
EG/BE=AG/AB
EG/BG=EG/(EG+BE)=AG/(AG+AB)
EG=AG*BG/(AG+AB)..........................................................(2)
(1)/(2)得，
FG/EG=CG*(AG+AB)/(BG*BG)...........................................(3)
BF平分∠ABC，
AG/CG=AB/BC
AG/AC=AG/(AG+CG)=AB/(AB+BC)
AG=AC*AB/(AB+BC)=AC*(√2-1)..........................................(4)
CG=AC-AG=AC*(2-√2)........................................................(5)
所以，CG*(AG+AB)=AC*AC2*(√2-1)
BG*BG=AB*AB+AG*AG=AC*AC*(4-2√2)..............................(6)
把(4)、(5)\、(6)代入(3)得，
FG/EG=1/√2
即，EG=FG*√2