Giải thích các bước giải:
a.Ta có : $AD\perp BC, BE\perp AC\to\widehat{ADC}=\widehat{BEC}=90^o$
$\to \Delta ADC\sim\Delta BEC(g.g)$
$\to \dfrac{CD}{CE}=\dfrac{CA}{CB}\to CD.CB=CA.CE$
b.Từ câu a $\to \dfrac{CE}{CB}=\dfrac{CD}{CA}\to \Delta CED\sim\Delta CBA(c.g.c)$
$\to \widehat{DEC}=\widehat{ABC}$
c.Ta có : $\widehat{BHD}=\widehat{BEC}=90^o$
$\to \Delta BDH\sim\Delta BEC(g.g)\to\dfrac{BH}{BC}=\dfrac{BD}{BE}\to BH.BE=BC.BD$
Tương tự $\to CH.CF=CD.CB$
$\to BH.BE=CH.CF=BC.BD+CD.CB=BC^2$