Giải thích các bước giải:
a.Ta có :
$AH\perp BC\to\widehat{BHA}=\widehat{BAC}$
$\to\Delta ABH\sim\Delta CBA(g.g)$
$\to\dfrac{AB}{CB}=\dfrac{BH}{AB}\to AB^2=BH.BC$
b.Ta có : $\widehat{CHA}=\widehat{CAB}=90^o\to\Delta AHC\sim\Delta BAC(g.g)$
$\to\dfrac{AC}{BC}=\dfrac{HC}{AC}\to AC^2=CH.BC$
c.Từ câu a $\to\widehat{BAH}=\widehat{ACB}=\widehat{ACH}$
Mà $\widehat{AHB}=\widehat{AHC}=90^o$
$\to\Delta ABH\sim\Delta CAH(g.g)$
$\to\dfrac{AH}{CH}=\dfrac{BH}{AH}\to AH^2=BH.CH$
d.Vì $AB\perp AC, AH\perp BC\to AH.BC=AB.AC(=2S_{ABC})$
