Giải thích các bước giải:
a.Ta có $\Delta ABC$ vuông tại $A$
$\to BC^2=AB^2+AC^2=289$
$\to BC=17$
Mà $AH\perp BC\to AH\cdot BC=AB\cdot AC(=2S_{ABC})$
$\to AH=\dfrac{AB\cdot AC}{BC}$
$\to AH=\dfrac{120}{17}$
$\to BH=\sqrt{AB^2-AH^2}=\dfrac{64}{17}$
$\to CH=BC-BH=\dfrac{225}{17}$
d.Ta có:
$\widehat{AHB}=\widehat{AHC}=90^o,\widehat{BAH}=90^o-\widehat{HAC}=\widehat{ACH}$
$\to \Delta AHB\sim\Delta CHA(g.g)$
$\to \dfrac{AH}{CH}=\dfrac{HB}{HA}$
$\to AH^2=HB.HC$
Lại có:
$\widehat{BHE}=\widehat{FHC},\widehat{BEH}=90^o-\widehat{EBH}=90^o-\widehat{DBC}=\widehat{BCD}=\widehat{HCF}$
$\to \Delta BHE\sim\Delta FHC(g.g)$
$\to \dfrac{BH}{FH}=\dfrac{HE}{HC}$
$\to HB.HC=HE.HF$
$\to AH^2=HB.HC$