Giải thích các bước giải:
1.Ta có:
$AB\perp BC\to BC^2 =AB^2+AC^2\to AC^2=BC^2-AB^2=16\to AC=4$
Mà $AK\perp BC\to AK.BC=AB.AC(=2S_{ABC})$
$\to AK=\dfrac{AB.AC}{BC}=\dfrac{12}{5}$
$\to BK^2=AB^2-AK^2=3^2-(\dfrac{12}{5})^2=\dfrac{81}{25}$
$\to BK=\dfrac95$
b.Ta có:
$\cos C=\dfrac{AC}{BC}=\dfrac45$
$\sin C=\dfrac{AB}{BC}=\dfrac35$
$\tan C=\dfrac{\sin C}{\cos C}=\dfrac34$
$\to H=5(\dfrac45+\dfrac35)-2\sqrt{1-\dfrac34}=6$
2.Ta có: $AB\perp AC, AK\perp BC\to \widehat{AKB}=\widehat{BAC}=90^o$
Mà $\widehat{ABK}=\widehat{ABC}$
$\to\Delta ABC\sim\Delta KBA(g.g)$
$\to\dfrac{AB}{KB}=\dfrac{AC}{AK}$
Lại có:
$BA=BD\to \Delta BAD$ cân tại $B\to\widehat{BAD}=\widehat{BDA}$
$\widehat{KAD}=90^o-\widehat{ADK}=90^o-\widehat{BAD}=\widehat{DAC}$
$\to AD$ là phân giác $\widehat{KAC}$
$\to\dfrac{DC}{DK}=\dfrac{AC}{AK}$
$\to \dfrac{DC}{DK}=\dfrac{AB}{KB}$
$\to \dfrac{KD}{KB}=\dfrac{DC}{AB}=\dfrac{DC}{BD}$ vì $BA=BD$
