cho tam giác ABC nhọn có trực tâm H cmr: tanA ×vectoHA +tanB × vectoHB +tanC × vectoHC = vecto 0

\[\begin{array}{l} \tan A.\overrightarrow {HA} + \tan B.\overrightarrow {HB} + \tan C.\overrightarrow {HC} = \overrightarrow 0 \\ \Leftrightarrow \overrightarrow {HC} = - \frac{{\tan A}}{{\tan C}}\overrightarrow {HA} - \frac{{\tan B}}{{\tan C}}\overrightarrow {HB} \\ Ve\,hinh\,binh\,hanh\,HB'CA'\,ta\,co:\\ \left. \begin{array}{l} \frac{{HB'}}{{HB}} = \frac{{DC}}{{DB}}\\ \frac{{\tan B}}{{\tan C}} = \frac{{AD}}{{BD}}:\frac{{AD}}{{DC}} = \frac{{DC}}{{DB}} \end{array} \right\} \Rightarrow \frac{{HB'}}{{HB}} = \frac{{\tan B}}{{\tan C}}\\ \Rightarrow \overrightarrow {HB'} = - \frac{{\tan B}}{{\tan C}}\overrightarrow {HB} \left( {do\,\overrightarrow {HB} ,\overrightarrow {HB'} \,nguoc\,huong} \right)\\ Tuong\,tu\,\overrightarrow {HA'} = - \frac{{\tan A}}{{\tan C}}\overrightarrow {HA} \\ \Rightarrow \overrightarrow {HB'} + \overrightarrow {HA'} = - \frac{{\tan B}}{{\tan C}}\overrightarrow {HB} - \frac{{\tan A}}{{\tan C}}\overrightarrow {HA} \\ Ma\,\overrightarrow {HB'} + \overrightarrow {HA'} = \overrightarrow {HC} \,nen\,\overrightarrow {HC} = - \frac{{\tan B}}{{\tan C}}\overrightarrow {HB} - \frac{{\tan A}}{{\tan C}}\overrightarrow {HA} \\ \Rightarrow dpcm \end{array}\]