[tex]m( \ \textless \ B)=60 \textdegree \Rightarrow m(\ \textless \ BAD)=30 \textdegree \Rightarrow BD=\frac{AB}{2} \Rightarrow \frac{DB}{AB}= \frac{1}{2} . [/tex]
[AE]-mediana in ΔABC-dreptunghic in A ⇒ [tex]AE= \frac{BC}{2}=BE=CE .[/tex]
[tex]AE=BE~si~m(\ \textless \ B)=60 \textdegree \Rightarrow \Delta ABE - echilateral \Rightarrow D-mijlocul ~ \\ lui~[BC] \Rightarrow DE= \frac{BE}{2}= \frac{BC}{4}. \\ \\ \frac{DE}{EC}= \frac{ \frac{BC}{4} }{ \frac{BC}{2} } = \frac{1}{2}. \\ \\ Observam~ca~ \frac{DB}{AB}= \frac{DE}{EC}= \frac{1}{2}.~~~(Q.E.D.) [/tex]
