E.622. Fie triunghiul ABCABCABC cu A^=60°,\widehat A=60\degree,A=60°, bisectoarea BE=5BE=5BE=5 cm, E∈ACE \in ACE∈AC și bisectoarea CD=10CD=10CD=10 cm, D∈AB.D \in AB.D∈AB. Să se demonstreze că triunghiul ABCABCABC este dreptunghic.
Fie FFF mijlocul lui DC.DC.DC. EIC^=B^2+C^2=60°; A^=60°⇒ADIE\widehat{EIC}=\dfrac{\widehat{B}}{2} + \dfrac{\widehat{C}}{2} = 60\degree;~ \widehat{A}=60\degree \Rightarrow ADIEEIC=2B+2C=60°; A=60°⇒ADIE inscriptibil ⇒D1^=E1^=30°.\Rightarrow \boxed{\widehat{D_1} = \widehat{E_1}=30\degree.}⇒D1=E1=30°. BE=FD, IE=ID⇒IB=IF⇒B1^=F1^=EIC^2=30°.BE=FD,~ IE=ID \Rightarrow IB=IF \Rightarrow \boxed{\widehat{B_1} = \widehat{F_1}=\dfrac{\widehat{EIC}}{2} =30\degree.}BE=FD, IE=ID⇒IB=IF⇒B1=F1=2EIC=30°.
Construim DG∥AC⇒D3^=A^=60°DG \parallel AC \Rightarrow \boxed{\widehat{D_3}=\widehat{A} =60\degree}DG∥AC⇒D3=A=60° și D2^=C1^=C2^⇒GFD^=90°⇒F2^=60°.\widehat{D_2} = \widehat{C_1} = \widehat{C_2} \Rightarrow \widehat{GFD}=90\degree \Rightarrow \boxed{\widehat{F_2}=60\degree}.D2=C1=C2⇒GFD=90°⇒F2=60°. Deci DBGFDBGFDBGF inscriptibil ⇒B2^=D2^=C2^⇒BF=FC=FD⇒B^=90°.\Rightarrow \widehat{B_2} = \widehat{D_2} = \widehat{C_2} \Rightarrow BF=FC=FD \Rightarrow \boxed{\widehat{B}=90\degree}.⇒B2=D2=C2⇒BF=FC=FD⇒B=90°.
