Giả sử tam giác đã cho là △ABC có a=AB;b=BC;c=CA
Và \(\widehat{A}=\dfrac{\widehat{B}+\widehat{C}}{2}\Leftrightarrow2\widehat{A}=\widehat{B}+\widehat{C}\Leftrightarrow3\widehat{A}=\widehat{A}+\widehat{B}+\widehat{C}=180^0\Leftrightarrow\widehat{A}=60^0\left(1\right)\)
Ta có \(\sqrt{a+b-c}=\sqrt{a}+\sqrt{b}-\sqrt{c}\Leftrightarrow a+b-c=a+b+c+2\sqrt{ab}-2\sqrt{ac}-2\sqrt{bc}\Leftrightarrow2\sqrt{bc}+2\sqrt{ac}-2\sqrt{ab}-c=0\Leftrightarrow\sqrt{bc}+\sqrt{ac}-\sqrt{ab}-c=0\Leftrightarrow\sqrt{bc}-\sqrt{ab}-c+\sqrt{ac}=0\Leftrightarrow\sqrt{b}\left(\sqrt{c}-\sqrt{a}\right)-\sqrt{c}\left(\sqrt{c}-\sqrt{a}\right)=0\Leftrightarrow\left(\sqrt{c}-\sqrt{a}\right)\left(\sqrt{b}-\sqrt{c}\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{c}-\sqrt{a}=0\\\sqrt{b}-\sqrt{c}=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}a=c\\b=c\end{matrix}\right.\)\(\Rightarrow\)△ABC cân(2)
Từ (1),(2)\(\Rightarrow\)△ABC đều
