YLL討論網

#ed_op#DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#IMG alt="image file name: 2kacb46d2308.jpg" src="http://yll.loxa.edu.tw/phpBB2/richedit/upload/2kacb46d2308.jpg" border=0#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#1.由題知 BF=BC#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#2.作角PCB=60度  =>三角形PBC為正三角形 => BP=PC=BC   又BF=BC#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#   =>BP=BF#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#3.連PF    因為BP=BF => 角BPF=角BFP=80度#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#4.在AC上取CK=PC    角PKC=角KPC=80度#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#5.連KF   易知三角形FPC全等於FKC (SAS)  =>KF=PF  又角PFK=60度  #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#     =>三角形 PFK為正三角形 =>PK=KF=PF#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#6.角PKC=角KPC=80度      角PKC=角PAK+角APK  =>角APK =角PAK=40度#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#        =>AK=PK  又PK=KF=PF  =>AK=KF#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#7.角FKC=20度=角AFK+角KAF  =>角AFK=角KAF=10度  =>角BAG=30度#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#   =>角BGA=90度#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#8.所求角GFC=180度-角FGC-角FCG=180-90-70=20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##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##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##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

是20度,我做出来了

我覺得應該是20度,不過我無法證明,想請教各位高手!

三角形ABC中，角BAC=40度，角ABC=60度，D、E分別在邊AC、AB上，使得角CBD=40度，角BCE=70度，BD和CE交於F，AF和BC交於G，試求角GFC的度數。