Giải thích các bước giải:
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 $\widehat{AEH}=\widehat{AHC}=90^o,\widehat{HAE}=\widehat{HAC}$
$\to \Delta AHE\sim\Delta ACH(g.g)$
$\to\dfrac{AH}{AC}=\dfrac{AE}{AH}$
$\to AH^2=AE.AC$
$\to AE.AC=HB.HC(đpcm)$
$\to AE=HB.\dfrac{HC}{AC}$
$\to AE.BC=HB.\dfrac{HC.BC}{AC}$
Ta có $\widehat{CHA}=\widehat{CAB}=90^o,\widehat{ACH}=\widehat{ACB}$
$\to \Delta CHA\sim\Delta CAB(g.g)$
$\to\dfrac{CH}{CA}=\dfrac{CA}{CB}$
$\to \dfrac{BC.CH}{CA}=CA$
$\to AE.BC=HB.AC$
