Giải thích các bước giải:
a.Ta có : $ME//AD\to \dfrac{ME}{AD}=\dfrac{MB}{BD}$
$MF//AD\to \dfrac{MF}{AD}=\dfrac{CM}{CD}$
$\to \dfrac{ME}{AD}+\dfrac{MF}{AD}=\dfrac{MB}{BD}+\dfrac{CM}{CD}$
Vì D là trung điểm BC$\to DB=DC=\dfrac12BC$
$\to \dfrac{ME}{AD}+\dfrac{MF}{AD}=\dfrac{MB}{\dfrac12BC}+\dfrac{CM}{\dfrac12BC}$
$\to \dfrac{ME+MF}{AD}=\dfrac{2MB}{BC}+\dfrac{2CM}{BC}$
$\to \dfrac{ME+MF}{AD}=\dfrac{2MB+2CM}{BC}$
$\to \dfrac{ME+MF}{AD}=\dfrac{2(MB+CM)}{BC}$
$\to \dfrac{ME+MF}{AD}=\dfrac{2BC}{BC}$
$\to \dfrac{ME+MF}{AD}=2$
$\to ME+MF=2AD$
b.Vì I là trung điểm EF
$\to MI=ME+EI=ME+\dfrac12EF=\dfrac12(2ME+EF)=\dfrac12(ME+(ME+EF))=\dfrac12(ME+MF)=\dfrac12\cdot 2AD=AD$
Mà $AD//MF\to AD//MI\to ADMI$ là hình bình hành