Ta có: \(CN=\frac12AN\)
=>\(S_{BNC}=\frac12\cdot S_{BNA};S_{ENC}=\frac12\cdot S_{ENA}\)
=>\(S_{BNC}-S_{ENC}=\frac12\cdot\left(S_{BNA}-S_{ENA}\right)\)
=>\(S_{BEC}=\frac12\cdot S_{BEA}\)
Ta có: \(BM=\frac13AM\)
=>\(S_{CMB}=\frac13\cdot S_{CMA};S_{EMB}=\frac13\cdot S_{EMA}\)
=>\(S_{CMB}-S_{EMB}=\frac13\cdot\left(S_{CMA}-S_{EMA}\right)\)
=>\(S_{CEB}=\frac13\cdot S_{CEA}\)
=>\(\frac12\cdot S_{AEB}=\frac13\cdot S_{AEC}\)
=>\(S_{AEB}=\frac23\cdot S_{AEC}\)
Vì F nằm giữa B và C
nên \(\frac{S_{AFB}}{S_{AFC}}=\frac{FB}{FC};\frac{S_{EFB}}{S_{EFC}}=\frac{FB}{FC}\)
=>\(\frac{S_{AFB}-S_{EFB}}{S_{AFC}-S_{EFC}}=\frac{FB}{FC}\)
=>\(\frac{S_{AEB}}{S_{AEC}}=\frac{FB}{FC}\)
=>\(\frac{FB}{FC}=\frac23\)
=>\(\frac{CF}{FB}=\frac32\)
