Giải thích các bước giải:
a.Ta có:
$BM=3MC$
$\to BM+3BM=3MC+3BM$
$\to 4BM=3(MC+BM)$
$\to 4BM=3BC$
$\to BM=\dfrac34BC$
$\to S_{ABM}=\dfrac34S_{ABC}=\dfrac34\cdot 160=120$
b.Ta có:
$AN=3NB\to AN+NB=3NB+NB\to AB=4BN$
$\to BN=\dfrac14AB$
$\to S_{BNC}=\dfrac14S_{ABC}=\dfrac14\cdot 160=40$