Kẻ MK//BC
=>AM/MB=AK/KC=2
=>AK=2KC
=>AK=2/3AC
mà AN=1/2AC
nên AK/AN=4/3
=>AN/AK=3/4
=>\(S_{ANM}=\dfrac{3}{4}\cdot S_{AMK}\)
=>\(S_{AMK}=108\left(cm^2\right)\)
ΔABC có MK//BC
nên ΔAMK đồng dạng vơi ΔABC
=>\(\dfrac{S_{AMK}}{S_{ABC}}=\left(\dfrac{AM}{AB}\right)^2=\left(\dfrac{2}{3}\right)^2=\dfrac{4}{9}\)
=>\(S_{ABC}=108:\dfrac{4}{9}=27\cdot9=243\left(cm^2\right)\)