Giải thích các bước giải:
a.Ta có $\Delta ABC$ vuông tại $A, AH\perp BC$
$\to S_{ABC}=\dfrac12AB.AC=\dfrac12AH.BC$
$\to AH.BC=AB.AC$
b.Xét $\Delta AHB,\Delta ABC$ có:
Chung $\hat B$
$\widehat{AHB}=\widehat{BAC}=90^o$
$\to\Delta BAH\sim\Delta BCA(g.g)$
$\to \dfrac{BA}{BC}=\dfrac{BH}{BA}$
$\to AB^2=BH.BC$
c.Xét $\Delta AHB,\Delta HAC$ có:
$\widehat{AHB}=\widehat{AHC}=90^o$
$\widehat{BAH}=90^o-\widehat{HAC}=\widehat{HCA}$
$\to\Delta AHB\sim\Delta CHA(g.g)$
$\to \dfrac{AH}{CH}=\dfrac{HB}{HA}$
$\to HA^2=HB.HC$
d.Ta có $\Delta ABC$ vuông tại $A$
$\to AB^2+AC^2=BC^2$
$\to \dfrac{AB^2+AC^2}{AB^2.AC^2}=\dfrac{BC^2}{AB^2.AC^2}$
$\to \dfrac{AB^2+AC^2}{AB^2.AC^2}=\dfrac{BC^2}{(AB.AC)^2}$
$\to \dfrac{AB^2+AC^2}{AB^2.AC^2}=\dfrac{BC^2}{(AH.BC)^2}$
$\to \dfrac{AB^2+AC^2}{AB^2.AC^2}=\dfrac{BC^2}{AH^2.BC^2}$
$\to \dfrac{1}{AB^2}+\dfrac{1}{AC^2}=\dfrac{1}{AH^2}$
$\to đpcm$
