由 jackhoung 於 星期二 五月 16, 2006 6:25 pm
#ed_op#DIV#ed_cl#因AD為圓O1,O2之公弦#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#所以O1O2垂直平分AD#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#即直線O1O2為AD之中垂線#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#AH之垂直平分線交AD之垂直平分線於M#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#M為三角形ADH之外心#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#作O1P垂直BD於P#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#O2Q垂直CD於Q#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#MR垂直DH於R#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#則O1P平行O2Q平行MR#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#又PR=1/2*BD+1/2*DH=1/2*BH#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#QR=1/2*CD-1/2*DH=1/2CH#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#PQ=QR#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#O1M=O2M#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#M為O1O2之中點#ed_op#/DIV#ed_cl#