Ta có:
$\,\,\,\,{{S}_{ABC}}={{S}_{A\text{D}B}}+{{S}_{A\text{D}C}}$
$\Leftrightarrow \frac{1}{2}bc.\operatorname{Sin}\widehat{\text{A}}=\frac{1}{2}A\text{D}.c.\operatorname{Sin}\frac{\widehat{A}}{2}+\frac{1}{2}A\text{D}.b.\operatorname{Sin}\frac{\widehat{A}}{2}$
$\Leftrightarrow 2bc.\operatorname{Sin}\frac{\widehat{A}}{2}.Cos\frac{\widehat{A}}{2}=A\text{D}.\operatorname{Sin}\frac{\widehat{A}}{2}.\left( b+c \right)$
$\Leftrightarrow 2bc.Cos\frac{\widehat{A}}{2}=A\text{D}.\left( b+c \right)$
$\Leftrightarrow A\text{D}=\frac{2bc.Cos\frac{\widehat{A}}{2}}{b+c}$
Mà $\cos \widehat{A}=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$
$=\frac{{{\left( b+c \right)}^{2}}-{{a}^{2}}}{4bc}$
Vì 0o < $\frac{\widehat{A}}{2}$ < 180o
$\Rightarrow Cos\frac{\widehat{A}}{2}=\sqrt{\frac{{{\left( b+c \right)}^{2}}-{{a}^{2}}}{4bc}}$
$\Rightarrow A\text{D}=\frac{2bc.\sqrt{\frac{{{\left( b+c \right)}^{2}}-{{a}^{2}}}{4bc}}}{2\left( b+c \right)}=\frac{\sqrt{bc\left[ {{\left( b+c \right)}^{2}}-{{a}^{2}} \right]}}{2\left( b+c \right)}$