Cần chứng minh: \(\tan A.\overrightarrow {HA} + \tan B.\overrightarrow {HB} + \tan C.\overrightarrow {HC} = \overrightarrow 0 \)

⇔ \[\overrightarrow {HC} = - \frac{{\tan A}}{{\tan C}}.\overrightarrow {HA} - \frac{{\tan B}}{{\tan C}}.\overrightarrow {HB} \]

Vẽ hình bình hành HB’CA’ ta có:

\(\left\{ \begin{array}{l}\frac{{HB'}}{{HB}} = \frac{{DC}}{{DB}}\\\frac{{\tan B}}{{\tan C}} = \frac{{AD}}{{BD}}:\frac{{AD}}{{DC}} = \frac{{DC}}{{DB}}\end{array} \right.\)

Suy ra: \(\frac{{HB'}}{{HB}} = \frac{{\tan B}}{{\tan C}}\)

⇒ \[\overrightarrow {HB'} = - \frac{{\tan B}}{{\tan C}}.\overrightarrow {HB} \](do \[\overrightarrow {HB'} ;\overrightarrow {HB} \]ngược hướng)

Tương tự: \[\overrightarrow {HA'} = - \frac{{\tan A}}{{\tan C}}.\overrightarrow {HA} \]

⇒ \[\overrightarrow {HB'} + \overrightarrow {HA'} = - \frac{{\tan B}}{{\tan C}}.\overrightarrow {HB} - \frac{{\tan A}}{{\tan C}}.\overrightarrow {HA} \]

Mà: \[\overrightarrow {HB'} + \overrightarrow {HA'} = \overrightarrow {HC} \]

Nên: \[\overrightarrow {HC} = - \frac{{\tan A}}{{\tan C}}.\overrightarrow {HA} - \frac{{\tan B}}{{\tan C}}.\overrightarrow {HB} \]

Vậy \(\tan A.\overrightarrow {HA} + \tan B.\overrightarrow {HB} + \tan C.\overrightarrow {HC} = \overrightarrow 0 \).