关于蝴蝶定理的一道几何题四边形AXBY中,XA=XB,YA=YB,XY交AB于M,过M作线段CD和EF,C在AX上,D在

连接AE,AD,BF,BC
IM/AI*BJ/JM=S△MCF/S△ACF* S△BED/S△MED
=S△AXY/S△ACF * S△BED/S△BXY * S△MCF/S△MED (因为S△AXY=S△BXY)
=(AX*AY/AC*AF)*(BE*BD/BX*BY)*(MC*MF/ME*MD)
=(BE*BD/AC*AF) *(MC/MD)*(MF/ME) (因为XA=XB,YA=YB)
=(BE/AC)*(BD/AF)*(MC/MD)*(MF/ME)
=(S△EAB/S△CAB)*(S△DAB/S△FAB)*(S△CAB/S△DAB)*(S△FAB*S△EAB)
=1
IM/AI=JM/BJ
∵IM+AI=AM=BM=JM+BJ
∴MI=MJ