Giải thích các bước giải:
a.Ta có: $AC=AD+DC=8$
$\to \dfrac{AB}{AD}=\dfrac{4}{2}=\dfrac{8}{4}=\dfrac{AC}{AB}$
Mà $\widehat{BAD}=\widehat{BAC}$
$\to\Delta ABD\sim\Delta ACB(c.g.c)$
$\to\widehat{ABD}=\widehat{ACB}=20^o$
b.Từ câu a
$\to \dfrac{BD}{CB}=\dfrac{AB}{AC}=\dfrac12$
$\to BD=\dfrac12BC$
Mà $\widehat{ACB}=20^o$
$\to CA^2+CB^2-AB^2=2CA\cdot CB\cdot\cos20^o$
$\to 8^2+CB^2-4^2=2\cdot 8\cdot CB\cdot\cos20^o$
$\to CB^2-16CB\cos20^o+48=0$
$\to CB\approx 10.44$ hoặc $CB\approx 4.59$
$\to BD\approx 5.22$ hoặc $BD\approx 2.295$
