a: BM+MC=BC
=>\(BC=BM+2\times BM=3\times BM\)
=>\(BM=\frac13\times BC\)
=>\(S_{ABM}=\frac13\times S_{ABC}\) (1)
Ta có; BN+NA=BA
=>BA=2BN+BN=3BN
=>\(BN=\frac13\times BA\)
=>\(S_{BNC}=\frac13\times S_{BAC}\) (2)
Từ (1),(2) suy ra \(S_{BMA}=S_{BNC}\)
b: Ta có: \(S_{BNOM}+S_{ANO}=S_{ABM}\)
\(S_{BNOM}+S_{OMC}=S_{BNC}\)
mà \(S_{BMA}=S_{BNC}\)
nên \(S_{ANO}=S_{OMC}\)
