[tex]\it m(\hat B)=60^o\ \ \ (complementul\ lui\ 30^o)\\ \\ Fie\ B'\ simetricul \ lui\ B\ fa\c{\it t}\breve a\ de\ AC\ \Rightarrow AB'=AB=\dfrac{BB'}{2}\\ \\ \\ \left.\begin{aligned} \it AB=AB'\\ \\ AC=AC\ latur\breve a\ comun\breve a\\ \\ \it \widehat{CAB}=\widehat{CAB'}=90^o\end{aligned}\right\}\ \stackrel{CC}{\Longrightarrow}\ \Delta CAB\equiv \Delta CAB'} \Rightarrow\begin{cases} \it \widehat{B'CB}=60^o\\ \\ \it \widehat{CB'B}=\widehat{CBB'}=60^o\end{cases} \Rightarrow[/tex]
[tex]\it \Rightarrow \Delta CBB'-echilateral \Rightarrow BB'=BC \Rightarrow \dfrac{BB'}{2}=\dfrac{BC}{2} \Rightarrow AB=\dfrac{BC}{2}[/tex]
