[tex]\it Fie\ AC\cap BD=\{O\}\\ \\ m(\hat{A})=60^o \Rightarrow\Delta ABD-echilateral \Rightarrow BD=AB=AD=4\ cm\\ \\ OD=BD:2=4:2=2\ cm\\ \\ \Delta AOD-dreptunghic,\ m(\hat{O})=90^o \stackrel{T.Pitagora}{\Longrightarrow}\ AO^2=AD^2-OD^2=[/tex]
[tex]\it = 4^2-2^2=16-4=12 \Rightarrow AO=\sqrt{12}=\sqrt{4\cdot3}=2\sqrt3\ cm\\ \\ MA\perp (ABC),\ AO\perp BD, AO,\ BD \subset (ABC) \stackrel{T.3\perp}{\Longrightarrow}MO\perp BD \Rightarrow \\ \\ \Rightarrow d(M,\ BD)=MO\\ \\ \Delta MAO-dreptunghic,\ m(\hat{A})=90^o\stackrel{T.Pitagora}{\Longrightarrow}\ \ MO^2=AM^2+AO^2=\\ \\ =6^2+(2\sqrt3)^2=36+12=48\Rightarrow MO=\sqrt{48}=\sqrt{16\cdot3}=4\sqrt3\ cm[/tex]
