Giải thích các bước giải:
Xét $\Delta ABC$ có:
$\cos B=\dfrac{a^2+c^2-b^2}{2ac}$(định lý cos)
$\to \cos\widehat{ABD}=\dfrac{a^2+c^2-b^2}{2ac}$
Vì $AD$ là phân giác $\hat A\to \dfrac{DB}{DC}=\dfrac{AB}{AC}=\dfrac{c}{b}$
$\to \dfrac{BD}{BD+DC}=\dfrac{c}{c+b}$
$\to \dfrac{BD}{BC}=\dfrac{c}{b+c}$
$\to \dfrac{BD}{a}=\dfrac{c}{b+c}$
$\to DB=\dfrac{ac}{b+c}$
Xét $\Delta ABD$ có:
$\cos\widehat{ABD}=\dfrac{BA^2+BD^2-AD^2}{2BA\cdot BD}$
$\to \dfrac{a^2+c^2-b^2}{2ac}=\dfrac{c^2+(\dfrac{ac}{b+c})^2-AD^2}{2\cdot c\cdot \dfrac{ac}{b+c}}$
$\to c^2+(\dfrac{ac}{b+c})^2-AD^2=\dfrac{a^2+c^2-b^2}{2ac}\cdot 2\cdot c\cdot \dfrac{ac}{b+c}$
$\to AD^2=c^2+(\dfrac{ac}{b+c})^2-(\dfrac{a^2+c^2-b^2}{2ac}\cdot 2\cdot c\cdot \dfrac{ac}{b+c})$
$\to AD^2=\dfrac{bc(c^2+2bc+b^2-a^2)}{\left(c+b\right)^2}$
$\to AD^2=\dfrac{bc(b+c)^2-a^2bc}{\left(c+b\right)^2}$
$\to AD^2=bc-\dfrac{a^2bc}{(c+b)^2}$
$\to AD=\sqrt{bc-\dfrac{a^2bc}{(c+b)^2}}$
