|
|
Ta có:
$\cot A=\frac{\cos A}{\sin A}=\frac{\displaystyle\frac{b^2+c^2-a^2}{2bc}}{\displaystyle \frac{2S}{bc}}=\frac{b^2+c^2-a^2}{4S}$
Tương tự:
$\cot B=\frac{a^2+c^2-b^2}{4S},\cot C=\frac{a^2+b^2-c^2}{4S}$
Từ đó:
$
\frac{1}{3} .\frac{m^2_a+m^2_b+m^2_c}{\cot A+\cot B+\cot C}=\frac{1}{3}.\frac{\displaystyle \frac{2b^2+2c^2-a^2}{4}+
\frac{2a^2+2c^2-b^2}{4}+\frac{2a^2+2b^2-c^2}{4}}{\displaystyle\frac{b^2+c^2-a^2}{4S}+\frac{a^2+c^2-b^2}{4S}+\frac{a^2+b^2-c^2}{4S}}$
$=\frac{1}{3}.\frac{3S(a^2+b^2+c^2)}{(a^2+b^2+c^2)}=S$
|