Đáp án: $AD=\dfrac{bc}{b+c}\cdot\sqrt{2(1+\cos(\hat A))}$
Giải thích các bước giải:
Đặt $BC=a$
Vì $AD$ là phân giác góc $A$
$\to \dfrac{DB}{DC}=\dfrac{AB}{AC}=\dfrac{c}{b}$
$\to\dfrac{DB}{DB+DC}=\dfrac{c}{c+b}$
$\to\dfrac{DB}{BC}=\dfrac{c}{c+b}$
$\to\dfrac{DB}{a}=\dfrac{c}{c+b}$
$\to BD=\dfrac{ac}{b+c}$
Tương tự $CD=\dfrac{ab}{b+c}$
Kẻ $DE//AB, E\in AC$
$\to \dfrac{DE}{AB}=\dfrac{DC}{BC}=\dfrac{b}{b+c}$
$\to DE=\dfrac{AB\cdot b}{b+c}=\dfrac{bc}{b+c}$
Mà $\widehat{EDA}=\widehat{DAB}=\widehat{DAC}=\widehat{DAE}$
$\to \Delta EAD$ cân tại $E$
$\to EA=ED=\dfrac{bc}{b+c}$
Ta có : $\widehat{AED}=180^o-\widehat{A}$
$\to AD^2=AE^2+ED^2-2AE.DE\cos\widehat{AED}$
$\to AD^2=2AE^2-2AE^2\cos(180^o-\hat A)$
$\to AD^2=2AE^2(1-\cos(180^o-\hat A))$
$\to AD^2=2AE^2(1+\cos(\hat A))$
$\to AD=AE\sqrt{2(1+\cos(\hat A))}$
$\to AD=\dfrac{bc}{b+c}\cdot\sqrt{2(1+\cos(\hat A))}$
