Lời giải:
Ta có:
$\quad \dfrac12AH.BC = \dfrac12AB.AC\quad (=S_{ABC})$
$\Leftrightarrow AH.BC = AB.AC$
$\Leftrightarrow \dfrac{1}{AH} = \dfrac{BC}{AB.AC}$
$\Leftrightarrow \dfrac{1}{AH^2} = \dfrac{BC^2}{AB^2.AC^2}$
$\Leftrightarrow \dfrac{1}{AH^2} = \dfrac{AB^2 + AC^2}{AB^2.AC^2}$
$\Leftrightarrow \dfrac{1}{AH^2} = \dfrac{AB^2}{AB^2.AC^2} + \dfrac{AC^2}{AB^2.AC^2}$
$\Leftrightarrow \dfrac{1}{AH^2} = \dfrac{1}{AB^2} + \dfrac{1}{AC^2}$