Răspuns:
Explicație pas cu pas:
[tex]x=\vec{AB} \: \vec{AC} = AB\times AC\times cos (BAC)\\y=\vec{BC}\:\vec{BA} = BC\times BA\times cos(ABC)\\z=\vec{CA}\:\vec{CB} = CA\times CB\times cos(ACB)\\\\x = y = z \rightarrow\\AB\times AC\times cos (BAC)= BC\times BA\times cos(ABC)= CA\times CB\times cos(ACB)\\AB\times AC\times cos (BAC)= BC\times AB\times cos(ABC)\\AB\times AC\times cos (BAC)= AC\times BC\times cos(ACB) \rightarrow \\AB\times cos(BAC) = BC \times cos(ACB)[/tex]
[tex]\\AB\times AC\times cos (BAC)= BC\times AB\times cos(ABC)\rightarrow\\AC\times cos(BAC) = BC\times cos(ABC)\\[/tex]
[tex]BC\times BA\times cos(ABC)= CA\times CB\times cos(ACB)\rightarrow\\AB\times cos(ABC)=AC\times cos(ACB)[/tex]
[tex]cos(BAC) = \frac{BC}{AB}\times cos(ACB)\\cos(BAC) = \frac{BC}{AC}\times cos(ABC)\\\rightarrow \frac{cos(ACB)}{AB}=\frac{cos(ABC)}{AC}\\\rightarrow AC\times cos(ACB) = AB\times cos(ABC)[/tex]
Acum, fie D proiectia punctului A pe latura BC. Acum avem:
[tex]CD = BD \rightarrow D\:este\:mijlocul\:lui\:BC\rightarrow AD-mediana\:si\:inaltime\rightarrow ABC\:isoscel(AB = AC)[/tex]
[tex]Analog, AC\times cos(BAC) = BC\times cos(ABC)\\AB\times cos(ABC)=AC\times cos(ACB)\\\\\rightarrow \frac{cos(BAC)}{BC} = \frac{cos(ACB)}{AB}\\\rightarrow AB \times cos(BAC) = BC\times cos(ACB)[/tex]
Acum, fie E proiectia lui C pe AB
[tex]EA = EB \rightarrow E - mijlocul\:lui\:AB\\\rightarrow CE-mediana\:si\:inaltime\\\rightarrow ABC\: isoscel(CA = CB)[/tex]
Deci, toate laturile sunt congruente[tex]\rightarrow[/tex] ABC - isoscel
