a: Xét ΔABD có ME//BD
nên ME/BD=AE/AD
Xét ΔADC có NE//DC
nên NE/DC=AE/AD=ME/BD
mà BD=CD
nên NE=ME
b: AM=2/3AB
=>AM=2MB
=>AB=3MB
\(\Leftrightarrow S_{AEB}=3\cdot S_{MEB}=3\left(cm^2\right)\)
AM/AB=2/3
=>AE/AD=2/3
=>\(S_{AEB}=\dfrac{2}{3}\cdot S_{ADB}\)
=>\(S_{ABD}=3:\dfrac{2}{3}=4.5\left(cm^2\right)\)
=>\(S_{ABC}=9\left(cm^2\right)\)