[tex]\displaystyle\it\\\boxed{\textnormal{Medianele triunghiului se intersecteaza intr-un punct numit centrul }}\\ \boxed{\it \textnormal{de greutate, notat G, se gaseste la } \frac{1}{3} \textnormal{ de baza si la}~\frac{2}{3}~\textnormal{fata de varf}.}[/tex]
[tex]\displaystyle\\\boxed{\textnormal{Mediana unui triunghi imparte triunghiul in doua triunghiuri de arii egale.}}\\----------------------------------[/tex]
[tex]\displaystyle\it\\\textnormal{Cum BN si AM sunt mediane si se intersecteaza in G}\implies \textnormal{G este centrul}\\ \textnormal{de greutate al triunghiului ABC.}~(*)\\\textnormal{a)} \stackrel{(*)}\Longrightarrow AG=\frac{2}{3}AM=12cm.\\\stackrel{(*)}\Longrightarrow BG=\frac{2}{3}BN=16cm.\\\textnormal{Observam ca in triunghiul ABG avem }: AB^2=AG^2+BG^2 \stackrel{R.T.P}\Longrightarrow \\\textnormal{triunghiul ABG este dreptunghic.}[/tex]
[tex]\displaystyle\it\\\textnormal{b) BN este mediana} \Longleftrightarrow \mathcal{A}_{ABN}=\mathcal{A}_{BNC}=\frac{1}{2}\mathcal{A}_{ABC} \Longleftrightarrow \mathcal{A}_{ABC}=2\mathcal{A}_{ABN}\\\Longleftrightarrow \mathcal{A}_{ABC}=2\left(\frac{BN\cdot d(A,BN)}{2}\right) \stackrel{a)}\Longrightarrow \mathcal{A}_{ABC}=BN\cdot AG=24\cdot 12=288cm^2.\\[/tex]
