Question

把一张长方形纸的B角任意折向形内折痕为EF再把CF折叠使CF与FB‘在同一条直线上试问两条折痕FE与FG的夹角

Answer 1

∠EFB=∠EFB',∠GFC=∠GFC'

∠EFB+∠EFB'+∠GFC+∠GFC'=180

∠EFB'+∠GFC'=180/2=90

FE与FG的夹角=90