Question

Chứng minh tam giác ABC thỏa mãn \(\left\{{}\begin{matrix}a^2=b^2+c^2-bc\\b^2=a^2+c^2-ac\end{matrix}\right.\)

thì là tam giác đều 

 

Answer 1

\(a^2=b^2+c^2-bc\Rightarrow bc=b^2+c^2-a^2\)

\(\Rightarrow cosA=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{bc}{2bc}=\dfrac{1}{2}\Rightarrow A=60^0\)

Tương tự: \(ac=a^2+c^2-b^2\Rightarrow cosB=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{1}{2}\Rightarrow B=60^0\)

\(\Rightarrow C=180^0-\left(A+B\right)=60^0\)

\(\Rightarrow A=B=C=60^0\Rightarrow\Delta ABC\) đều