YLL討論網

#ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#在三角形ABC中 #ed_op#BR#ed_cl#AB=AC #ed_op#BR#ed_cl#D、E分別是AB、AC的中點 #ed_op#BR#ed_cl#P是DE上任意一點 #ed_op#BR#ed_cl#直線BP、CP分別交AC、AB於G、F #ed_op#BR#ed_cl#求證：1/BF+1/CG為定值 #ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl##ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#因D、E分別是AB、AC的中點 ，所以BD=EC#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl##ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#DE//BC#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#DF/BF=DP/BC#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#BD/BF=(BC-DP)/BC-----(1)#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl##ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#GE/GC=PE/BC#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#EC/GC=(BC-PE)/BC-----(2)#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl##ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#(1)，(2)結合#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#BD(1/BF+1/CG)=(2BC-DE)/BC#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#      (1/BF+1/CG)=(2BC-DE)/(BC*BD)#ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#      #ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl##ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/SPAN#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl#因DE=BC/2，BD=AB/2，BC都是定值，所以(1/BF+1/CG)為定值。  #ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/SPAN#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl##ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SPAN class=postbody#ed_cl##ed_op#FONT size=2#ed_cl##ed_op#/FONT#ed_cl##ed_op#/SPAN#ed_cl# #ed_op#/DIV#ed_cl#

在三角形ABC中
AB=AC
D、E分別是AB、AC的中點
P是DE上任意一點
直線BP、CP分別交AC、AB於G、F
求證：1/BF+1/CG為定值