ΔFEC și ΔBDE isoscele ⇒∡FEC=∡FCE, ∡BED=∡DBE
[tex]DE \perp EF \Rightarrow \Delta DE F \ dreptunghic \\ \Rightarrow \measuredangle DE F=90^{\circ}\Rightarrow \measuredangle BED+\measuredangle FEC=90^{\circ}\\ \Rightarrow DBE+\measuredangle ECF=90^{\circ} \Rightarrow \measuredangle B+\measuredangle C=90^{\circ}\\ \Rightarrow \measuredangle A=90^{\circ} \Rightarrow \tt \Delta ABC \ dreptunghic[/tex]
