Giải thích các bước giải:
a.Ta có : $DM//AC\to\widehat{DMB}=\widehat{ACB}\to\Delta BDM\sim\Delta BAC(g.g)$
$\to\dfrac{BD}{BA}=\dfrac{DM}{AC}=\dfrac{BM}{BC}=\dfrac{1}{3}$
Vì $\dfrac{MB}{MC}=\dfrac12\to\dfrac{MB}{MB+MC}=\dfrac1{1+2}\to\dfrac{BM}{BC}=\dfrac13$
b.Từ câu a
$\to \dfrac{BD}{BA}=\dfrac{DM}{AC}=\dfrac{BM}{BC}=\dfrac{1}{3}$
$\to \dfrac{BD+DM+BM}{AB+AC+BC}=\dfrac13$
$\to \dfrac{P_{BDM}}{P_{ABC}}=\dfrac13$
$\to P_{BDM}=\dfrac13P_{ABC}=8$