+)
$\Delta A'AB\bot A:A'B^2=A'A^2+AB^2=(2a\sqrt5)^2+a^2=21a^2$
$\Delta ABC$ dùng định lý cosin:
$BC^2=AB^2+AC^2-2.AB.AC.\cos\widehat{BAC}=a^2+(2a)^2-2.a.2a.\cos120^o=7a^2$
$\Delta MBC\bot C:BM^2=BC^2+MC^2=7a^2+(a\sqrt5)^2=12a^2$
$\Delta A'C'M\bot C':A'M^2=A'C'^2+C'M^2=(2a)^2+(a\sqrt5)^2=9a^2$
$\Delta A'MB$ có: $A'M^2+BM^2=9a^2+12a^2=21a^2=A'B^2$
Theo định lý Pitago đảo suy ra $\Delta A'MB\bot M\Rightarrow MA'\bot MB$
+)
Ta có: $V_{MA'AB}=V_{MA'B'B}$
$V_{MA'B'C'}=V_{MABC}$
$V_{ABCA'B'C'}=V_{MA'AB}+V_{MA'B'B}+V_{MA'B'C'}+V_{MABC}=2(V_{MA'AB}+V_{MABC})$
$V_{ABCA'B'C'}=A'A.S_{ABC}$
$V_{MABC}=\dfrac13\dfrac{A'A}2.S_{ABC}$
$\Rightarrow V_{MA'AB}=A'A.S_{ABC}(\dfrac{1}2-\dfrac{1}6)$
$=2a\sqrt5.\dfrac12.a.2a.\sin120^o.\dfrac13=a^3\sqrt{15}$
$V_{MA'AB}=V_{AA'BM}=\dfrac13.d(A,(A'BM)).S_{A'BM}$
$=\dfrac13.d(A,(A'BM)).\dfrac12.\sqrt{12a^2}.\sqrt{9a^2}=d.a^2\sqrt3=a^3\sqrt{15}$
$\Rightarrow d=a\sqrt5$.
