Ta có: $S_{ABC} = S_{ABD} + S_{ACD}$
$\Leftrightarrow \dfrac{1}{2}AB.AC.sinA = \dfrac{1}{2}AB.AD.sin\widehat{BAD} + \dfrac{1}{2}AC.AD.sin\widehat{CAD}$
$\Leftrightarrow 2bc.sin\dfrac{A}{2}.cos\dfrac{A}{2} = c.AD.sin\dfrac{A}{2} + b.AD.sin\dfrac{A}{2}$
$\Leftrightarrow 2bc.sin\dfrac{A}{2}.cos\dfrac{A}{2} = AD.sin\dfrac{A}{2}(b + c)$
$\Leftrightarrow AD = \dfrac{2bc.cos\dfrac{A}{2}}{b + c}$