[tex]\it Vom\ nota\ AB=9a,\ cu\ a\ \in\mathbb{Q}.\\ \\ Conform\ enun\c{\it t}ului,\ vom\ avea:\ \begin{cases}\it BE=\dfrac{9a}{3}=3a\\ \\ \it EC=9a-3a=6a\\ \\ \it FC=\dfrac{2\cdot9a}{9}=2a\end{cases}[/tex]
Comparăm triunghiurile ABE și ECF :
[tex]\it \left.\begin{aligned}\dfrac{AB}{EC}= \dfrac{9a}{6a} = \dfrac{3}{2}\\ \\ \\ \dfrac{BE}{CF}= \dfrac{3a}{2a}= \dfrac{3}{2}\\ \\ \\ \widehat{ABE} =\widehat{ECF} = 90^o\end{aligned}\right\} \stackrel{LUL}{\Longrightarrow}\ \Delta ABE\sim \Delta ECF \Rightarrow \widehat{BEA} =\widehat{CFE} \ \ \ \ \ (1)[/tex]
[tex]\it Din\ \Delta ECF\ \Rightarrow\ \widehat{CFE} =90^0-\widehat{FEC}\ \ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow\ \widehat{BEA} =90^o-\widehat{FEC} \Rightarrow \widehat{BEA} +\widehat{FEC}=90^o[/tex]
[tex]\it In\ punctul\ E\ avem:\\ \\ \widehat{AEF} =180^o -(\widehat{BEA} +\widehat{FEC})=180^o-90^o=90^o \Rightarrow\\ \\ \Rightarrow \Delta AEF\ este\ dreptunghic,\ \widehat{AEF} =90^o[/tex]
