由 jackhoung 於 星期五 七月 28, 2006 12:58 pm
#ed_op#DIV#ed_cl#連接BE,BC,BF,BD#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#則角CEB=角ADB#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#角DFB=角ACB#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#又BC=BF#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#ΔCEB全等於ΔFDB#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#DIV#ed_cl#所以CE=DF#ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl#