Giải thích các bước giải:
a.Ta có $BB'//d, CC'//d$
$\to BB'//CC'$
$\to \dfrac{DB'}{DC'}=\dfrac{DB}{DC}=1$ vì $D$ là trung điểm $BC$
$\to DB'=DC'$
b.Ta có: BB'//d$
$\to \dfrac{AB}{AM}=\dfrac{AB'}{AG}$
Tương tự $\dfrac{AC}{AN}=\dfrac{AC'}{AG}$
$\to \dfrac{AB}{AM}+\dfrac{AC}{AN}=\dfrac{AB'}{AG}+\dfrac{AC'}{AG}$
$\to \dfrac{AB}{AM}+\dfrac{AC}{AN}=\dfrac{AB'+AC'}{AG}$
$\to \dfrac{AB}{AM}+\dfrac{AC}{AN}=\dfrac{AD-DB'+AD+DC'}{AG}$
$\to \dfrac{AB}{AM}+\dfrac{AC}{AN}=\dfrac{2AD}{AG}$ vì $DB'=DC'$
$\to \dfrac{AB}{AM}+\dfrac{AC}{AN}=3$ vì $G$ là trọng tâm $\Delta ABC\to \dfrac{AG}{AD}=\dfrac23$
Ta có $BB'//d$
$\to \dfrac{BM}{AM}=\dfrac{GB'}{GA}$
Tương tự $\dfrac{CN}{AN}=\dfrac{GC'}{GA}$
$\to \dfrac{BM}{AM}+\dfrac{CN}{AN}=\dfrac{GB'+GC'}{AG}$
$\to \dfrac{BM}{AM}+\dfrac{CN}{AN}=\dfrac{GD-D'B+GD+DC'}{AG}$
$\to \dfrac{BM}{AM}+\dfrac{CN}{AN}=\dfrac{2GD}{AG}$
$\to \dfrac{BM}{AM}+\dfrac{CN}{AN}=1$ vì $G$ là trọng tâm $\Delta ABC\to GA=2GD$
