由 galaxylee 於 星期四 四月 19, 2007 9:20 pm
#ed_op#DIV#ed_cl##ed_op#FONT size=2#ed_cl##ed_op#IMG alt="image file name: 2k632b92cd68.gif" src="http://yll.loxa.edu.tw/phpBB2/richedit/upload/2k632b92cd68.gif" border=0#ed_cl##ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT size=2#ed_cl#證明:用反證法,假設AB≠AC,不妨設AB>AC #ed_op#BR#ed_cl#(1)∠ACB>ABC,因此∠BCF=∠FCE=1/2∠ACB>1/2∠ABC=∠CBE=∠EBF #ed_op#BR#ed_cl#(2)在△BCF與△CBE中,因為BC=BC,BE=CF,∠BCF>∠CBE,所以BF>CE #ed_op#BR#ed_cl#(3)作平行四邊形BEGF,則∠EBF=∠FGE,EG=BF,FG=BE=CF,連接CG,△FCG為等腰三角形,所以∠FCG=∠FGC #ed_op#BR#ed_cl#(4)因為∠FCE>∠FGE,所以∠ECG<∠EGC,故CE>EG=BF,顯然和(2)矛盾 #ed_op#BR#ed_cl#所以AB=AC #ed_op#/FONT#ed_cl##ed_op#BR#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#