Giải thích các bước giải:
a.Vì $AK=KI=IH\to AH=3AK, AH=\dfrac32AI$
$\to \dfrac{AK}{AH}=\dfrac13, \dfrac{AI}{AH}=\dfrac23$
Ta có $MN//BC$
$\to \dfrac{MN}{BC}=\dfrac{AM}{AB}=\dfrac{AK}{AH}=\dfrac13$
$\to MN=\dfrac13BC=0.5$
Ta có $EF//CB$
$\to \dfrac{EF}{BC}=\dfrac{AE}{AB}=\dfrac{AI}{AH}=\dfrac23$
$\to EF=\dfrac23BC=1$
b.Do $AH\perp BC\to AH\perp MN, AH\perp EF\to AK\perp MN, AI\perp EF$
Ta có $S_{ABC}=270$
$\to \dfrac12AH\cdot CB=270$
$\to AH=360$
$\to AK=KI=IH=\dfrac13AH=120$
$\to AI=2AK=240$
$\to S_{MNFE}=S_{AEF}-S_{AMN}$
$\to S_{MNFE}=\dfrac12AI\cdot EF-\dfrac12AK\cdot MN$
$\to S_{MNFE}=\dfrac12\cdot 240\cdot 1-\dfrac12\cdot 120\cdot 0.5$
$\to S_{MNFE}=90$