Chứng minh rằng $2$ đường trung tuyến $AA' , BB'$ vuông góc với
nhau khi và chỉ khi $\cot C = 2(\cot A+\cot B)       (1)$
Chứng minh: Gọi $2$ đường trung tuyến $AA' $ và $BB'$ vuông góc với nhau tại trọng tâm $G$ khi và chỉ khi $\Delta ABC$ vuông tại $G \Leftrightarrow AG^2+BG^2=AB^2$
Ta có : $ AG = \frac{2 }{ 3} AA' \Leftrightarrow 9AG^2= 4AA'^2 = 2b^2+2c^2-a^2 $
Tương tự,  $9BG^2 = 2a^2+2c^2-b^2$
                $AG^2+BG^2=AB^2 \Leftrightarrow 9AG^2+9BG^2=9AB^2$
          $\Leftrightarrow 2b^2+2c^2-a^2+2a^2+2c^2-b^2=9c^2$
          $\Leftrightarrow a^2+b^2=5c^2$
          $\Leftrightarrow c^2+2 ab \cos C = 5c^2$ ( Do định lý cô sin )
          $\Leftrightarrow ab \cos C =2c^2$
          $\Leftrightarrow \cos C=2 \frac{ c^2}{ ab} \Leftrightarrow \cos C=2 \frac{4R^2 \sin ^2 C }{ 4R^2 \sin A \sin B}  $
          $\Leftrightarrow \frac{ \cos C  }{ \sin C} = 2 \frac{ \sin (A+B)}{ \sin A \sin B} $
          $\Leftrightarrow \cot C=2(\cot A+\cot B)$
Vậy đẳng thức $(1)$ đúng.
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