Giải thích các bước giải:

Ta có:

$AB.HC=AC.HA$

$\to \dfrac{HC}{HA}=\dfrac{AC}{AB}$

$\to \dfrac{HC^2}{HA^2}=\dfrac{AC^2}{AB^2}$

$\to \dfrac{HC^2+HA^2}{HA^2}=\dfrac{AC^2+AB^2}{AB^2}$

$\to \dfrac{AC^2}{AH^2}=\dfrac{AC^2+AB^2}{AB^2}=\dfrac{AC^2+AB^2-AC^2}{AB^2-AH^2}=\dfrac{AB^2}{HB^2}$

$\to \dfrac{AC}{AH}=\dfrac{AB}{HB}$

$\to AC.HB=AB.AH$

Do $AB.HC=AC.HA$

$\to AC.HB.AB.HC=AB.AH.AC.AH$

$\to HB.HC=HA^2$

$\to \dfrac{HB}{HA}=\dfrac{HA}{HC}$

Do $\widehat{AHB}=\widehat{AHC}(=90^o)$

$\to \Delta HAB\sim\Delta HCA(c.g.c)$
$\to \widehat{HAB}=\hat C$

$\to \widehat{BAC}=\widehat{HAB}+\widehat{HAC}=\hat C+\widehat{HAC}=90^o$