a)
[tex]\it BE=3DE \Rightarrow BE=3(BD-BE) \Rightarrow BE=3BD-3BE \Rightarrow 4BE=3BD \Rightarrow \\ \\ \Rightarrow \dfrac{BD}{BE}=\dfrac{4}{3}\ \ \ \ \ (*)\\ \\ \\ EM||DC \Rightarrow \Delta BCD\sim \Delta BME \Rightarrow \dfrac{CD}{EM}=\dfrac{BD}{BE}\ \stackrel{(*)}{\Longrightarrow} \dfrac{4\sqrt3}{EM}=\dfrac{4}{3} \Rightarrow EM=3\sqrt3[/tex]
b)
[tex]\it tg(ADB)=\dfrac{4\sqrt3}{4}=\sqrt3 \Rightarrow \widehat{ADB}=60^o\ \ \ \ (1)\\ \\ BE=3DE\Rightarrow BD-DE=3DE \Rightarrow BD=4DE\Rightarrow DE=\dfrac{BD} {4}\\ \\ Fie \ \ O\ -\ mijlocul\ lui \ BD \Rightarrow OD=\dfrac{BD}{2} \Rightarrow EO=DE=\dfrac{BD}{4} \Rightarrow \\ \\ \\ \Rightarrow AE\ -\ median\breve a\ \^{i}n\ \Delta OAD\ \ \ \ (2)[/tex]
[tex]\it OA=OD \ (jum\breve at\breve a\c{\it t}i\ de \ diagonale) \Rightarrow \Delta OAD \ -\ isoscel\ \ \ \ (3)\\ \\ (1),\ (3) \Rightarrow \Delta OAD\ -\ echilateral\ \ \ \ \ (4)\\ \\ (2),\ (4) \Rightarrow AE\ -\ \^{i}n\breve al\c{\it t}ime \Rightarrow AE\perp BD[/tex]
Deoarece perpendiculara din A pe BD este mai mică decât
oricare oblică din A pe BD, rezultă:
[tex]\it AE\leq AF,\ \ \forall \ F\in BD[/tex]
