Question

Cho tam giác ABC. Chứng minh tam giác ABC cân nếu:                         \(4m^2_a\) = b(b + 4c.cos A)

Answer 1

\(4m_a^2=b\left(b+4c.\cos A\right)=b^2+4bc.\cos A\Rightarrow m_a^2=\dfrac{b^2+4bc.cosA}{4}\)

\(m_a^2=\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}\)

\(\Rightarrow\dfrac{b^2+4bc.cosA}{4}=\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}\)

\(\Leftrightarrow b^2+4bc.cosA=2b^2+2c^2-a^2\)

\(\Leftrightarrow b^2+2c^2-a^2=4bc.cosA\)

\(\Leftrightarrow b^2+2c^2-a^2=4bc.\dfrac{b^2+c^2-a^2}{2bc}=2\left(b^2+c^2-a^2\right)\)

\(\Leftrightarrow b^2+2c^2-a^2=2b^2+2c^2-2a^2\)

\(\Leftrightarrow a^2=b^2\Leftrightarrow a=b\Rightarrow\left(đpcm\right)\)