Giải thích các bước giải:
a.Ta có $AH\perp BC\to\widehat{AHB}=\widehat{AHC}=90^o$
Mà $AH^2=HB.HC\to \dfrac{AH}{HB}=\dfrac{HC}{HA}$
$\to \Delta AHB\sim\Delta CHA(c.g.c)$
$\to\widehat{BAH}=\widehat{ACH}$
$\to\widehat{BAC}=\widehat{BAH}+\widehat{HAC}=\widehat{ACH}+\widehat{HAC}=90^o$
$\to \Delta ABC$ vuông tại $A$
b.Ta có $AH.BC=AB.AC$
$\to \dfrac12AB.AC=\dfrac12AH.BC$
$\to \dfrac12AB.AC=S_{ABC}$
Kẻ $BK\perp AC=K$
$\to S_{ABC}=\dfrac12BK.AC$
$\to \dfrac12BK.AC=\dfrac12AB.AC$
$\to BK=BA\to K\equiv A$
$\to BA\perp AC$
$\to \Delta ABC$ vuông tại $A$
c.Nếu $\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}$
$\to \dfrac{AH^2}{AB^2}+\dfrac{AH^2}{AC^2}=1$
$\to \sin^2\widehat{ABH}+\sin^2\widehat{ACH}=1$
$\to \sin^2\widehat{ABH}=1-\sin^2\widehat{ACH}$
$\to\sin^2\widehat{ABH}=\cos^2\widehat{ACH}$
$\to\sin^2\widehat{ABH}=\cos^2\widehat{ACH}$
$\to\sin^2\widehat{ABC}=\cos^2\widehat{ACB}$
$\to\sin^2\widehat{ABC}=\sin^2(90^o-\widehat{ACB})$
$\to \widehat{ABC}=90^o-\widehat{ACB}$
$\to\widehat{ABC}+\widehat{ACB}=90^o$
$\to \Delta ABC$ vuông tại $A$
