由 jackhoung 於 星期日 十月 01, 2006 7:55 pm
#ed_op#DIV#ed_cl#取BC中點I,連接EI,FI#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#則EI平行CD,FI平行AB#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#角IEF=角2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#角IFE=角1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#且EI=1/2*CD=1/2*AB=FI#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以角IEF=角IFE#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#即角1=角2#ed_op#/DIV#ed_cl#