a: AK=KI=IH
mà AK+KI+IH=AH
nên \(AK=KI=IH=\frac{AH}{3}\)
Xét ΔABH có MK//BH
nên \(\frac{AM}{AB}=\frac{AK}{AH}=\frac13\)
Xét ΔABC có MN//BC
nên \(\frac{MN}{BC}=\frac{AM}{AB}=\frac13\)
Xét ΔABH có EI//BH
nên \(\frac{AE}{AB}=\frac{AI}{HA}=\frac23\)
Xét ΔABC có EF//BC
nên \(\frac{EF}{BC}=\frac{AE}{AB}=\frac23\)
b: ΔABC có EF//BC
nên ΔAEF~ΔABC
=>\(\frac{S_{AEF}}{S_{ABC}}=\left(\frac{AE}{AB}\right)^2=\left(\frac23\right)^2=\frac49\)
=>\(S_{AEF}=\frac49\cdot S_{ABC}\)
Xét ΔAMN và ΔAEF có
\(\frac{AM}{AE}=\frac{AN}{AF}\) (=1/2)
\(\hat{MAN}\) chung
Do đó: ΔAMN~ΔAEF
=>\(\frac{S_{AMN}}{S_{AEF}}=\left(\frac{AM}{AE}\right)^2=\frac14\)
=>\(S_{AMN}=\frac14\cdot S_{AEF}\)
=>\(S_{AMN}=\frac14\cdot\frac49\cdot S_{ABC}=\frac19\cdot S_{ABC}\)
TA có: \(S_{AMN}+S_{MNFE}=S_{AEF}\)
=>\(S_{MNFE}=\frac13\cdot S_{ABC}=\frac13\cdot90=30\left(\operatorname{cm}^2\right)\)
