Ta có: $M,N,P$ là trung điểm các cạnh của $∆ABC$
$\to ∆MNP\sim ∆ABC \, (c.c.c)$
$\to \dfrac{S_{MNP}}{S_{ABC}}=\left(\dfrac{MN}{AB}\right)^2 =\dfrac{1}{4}$
$\to \dfrac{\dfrac{1}{3}S_{MNP}.d(S;(MNP))}{\dfrac{1}{3}S_{ABC}.d(S;(ABC))}=\dfrac{1}{4}$
$\to \dfrac{\dfrac{1}{3}S_{MNP}.d(S;(ABC))}{\dfrac{1}{3}S_{ABC}.d(S;(ABC))}=\dfrac{1}{4}$
$\to \dfrac{V_{S.MNP}}{V_{S.ABC}}=\dfrac{1}{4}$
hay $\dfrac{V'}{V}=\dfrac{1}{4}$
