Giải thích các bước giải:
a.Vì $KM//AC, HF//BC\to \Diamond OECM$ là hình bình hành$\to OM=CF$
Ta có :
$\dfrac{AK}{AB}+\dfrac{BE}{BC}+\dfrac{CF}{CA}$
$=\dfrac{CM}{CB}+\dfrac{BE}{BC}+\dfrac{CF}{CA}$
$=\dfrac{CM+BE}{BC}+\dfrac{CF}{CA}$
$=\dfrac{BC-EM}{BC}+\dfrac{CF}{CA}$
$=1-\dfrac{EM}{BC}+\dfrac{CF}{CA}$
$=1-\dfrac{EM}{BC}+\dfrac{OM}{CA}$
Lại có $OM//AC, OE//AB\to\Delta OEM\sim\Delta ABC(g.g)\to \dfrac{EM}{BC}=\dfrac{OM}{AC}$
$\to \dfrac{EM}{BC}-\dfrac{OM}{AC}=0$
$\to $$\dfrac{AK}{AB}+\dfrac{BE}{BC}+\dfrac{CF}{CA}=1$
b.Ta có :
$\dfrac{DE}{AB}+\dfrac{FH}{BC}+\dfrac{MK}{CA}$
$=\dfrac{EC}{BC}+\dfrac{AF}{AC}+\dfrac{BK}{AB}$
$=\dfrac{BC-BE}{BC}+\dfrac{AC-CF}{AC}+\dfrac{AB-AK}{AB}$
$=1-\dfrac{BE}{BC}+1-\dfrac{CF}{AC}+1-\dfrac{AK}{AB}$
$=3-(\dfrac{BE}{BC}+\dfrac{CF}{AC}+\dfrac{AK}{AB})$
$=3-1=2$