正方形ABCD的对角线AC上取一点E,使AE=CD,过点E做EF垂直AC,交BC于点F,求证AB=CE+CF
问题描述:
正方形ABCD的对角线AC上取一点E,使AE=CD,过点E做EF垂直AC,交BC于点F,求证AB=CE+CF
答
设AB = 1
AC = 根号2
AE = 1
CE = AC - AE = 根号2 -1
CF = 根号2 * CE = 2 - 根号2
AB = 1 = CE + CF = 1
得证