YLL討論網

#ed_op#DIV#ed_cl#這麼說我好像也是....下禮拜段考0.0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#這是我的做法#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#易證ΔAEC與ΔABD全等#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#以E為中心將ΔEAC順時針旋轉60度得三角形EBC'#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#連CC'，∵BDCC'為平行四邊形且ΔECC'為正三角形#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#∴∠FCC'=∠CFD=60#ed_op#/DIV#ed_cl#

連AF
則角EAC=60+角BAC=角BAD
AE=AB AD=AC 所以三角形EAC和三角形BAD全等.
所以角ADF=角ACF. 即A.F.C.D四點共圓.
角CFD=角CAD=60度.

(那是因為接近段考了阿...而且我們這些國三生也不能太常上網...雖然我照樣每天上YLL...= =)

#ed_op#DIV#ed_cl#有一個三角形ABC,以AC線段為一邊,向外作一正三角形ACD,以AB線段為一邊..#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#向外作一正三角形ABE,連接BD線段與CE線段交於一點F,請問角CFD角度為幾#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#度??#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#給大家娛樂的題目，最近上來的人好像比較少#ed_op#/DIV#ed_cl#