Giải thích các bước giải:

a.Ta có:

$\cos B=\dfrac{BA}{BC}=\dfrac12\to \hat B=60^o$

$AC=\sqrt{BC^2-AB^2}=2\sqrt3$

b.Vì $AB\perp AC,CK\perp BH$

$\to CH\perp DB, BH\perp DC$

$\to H$ là trực tâm $\Delta DBC$

$\to DH\perp CB$

$\to DH\perp BC$ tại $E$

$\to \widehat{DKH}=\widehat{BKC}(=90^o)$

Mà $\widehat{HDK}=\widehat{EDC}=90^o-\widehat{ECD}=90^o-\widehat{BCK}=\widehat{KBC}$

$\to \Delta DHK\sim\Delta BCK(g.g)$

$\to \dfrac{DK}{DH}=\dfrac{BK}{BC}=\sin\widehat{BCK}$

c.Xét $\Delta BAC,\Delta BED$ có:

Chung $\hat B$

$\hat A=\hat E(=90^o)$

$\to \Delta BAC\sim\Delta BED(g.g)$

$\to \dfrac{BA}{EB}=\dfrac{BC}{BD}$

$\to BE=\dfrac{BA.BD}{BC}=\dfrac{2\cdot 3}4=\dfrac32$

$\to S_{ABE}=\dfrac{BE}{BC}S_{ABC}=\dfrac{\dfrac32}{4}\cdot \dfrac12\cdot 2\cdot 2\sqrt3=\dfrac{3\sqrt3}4$