Ducem înălțimea AD, care este și mediană, deci :
CD = BD = 20:2 = 10
[tex]\it \Delta ADC-dr,\ \widehat D=90^o,\ \stackrel{T.P.}{\Longrightarrow}AD^2=AC^2-CD^2=26^2-10^2=\\ \\ =(26-10)(26+10)=16\cdot36=4^2\cdot6^2 \Rightarrow AD=4\cdot6=24\\ \\ \\ \mathcal{A}_{ABC}=\dfrac{BC\cdot AD}{2}=\dfrac{20\cdot24}{2}=10\cdot24=240[/tex]
[tex]\it \mathcal{A}=\dfrac{AC\cdot BE}{2} \Rightarrow 240=\dfrac{26\cdot BE}{2} \Rightarrow 240=13\cdot BE \Rightarrow BE=\dfrac{240}{13}\approx 18,5[/tex]
