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發表 aaddfg 於 星期五 七月 28, 2006 1:47 pm

圓內接四邊形外角=內對角(對角互補)
AEBD是圓內接四邊形
故角AEB+角ADB=180
角AEB+角CEB=180
所以角ADB=角CEB

[問題]我有問題

發表 tangpakchiu 於 星期五 七月 28, 2006 1:26 pm

#ed_op#DIV#ed_cl#想問問jackhoung,為什麼角CEB=角ADB???#ed_op#/DIV#ed_cl#

發表 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#

[問題]幾何題30

發表 tangpakchiu 於 星期五 七月 28, 2006 10:18 am

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