Giải thích các bước giải:
a.Vì $AD,BE,CF$ là đường cao $\Delta ABC$
$\to AD\perp BC, BE\perp AC, CF\perp AB$
$\to\widehat{AEB}=\widehat{AFC}=90^o$
$\to\Delta AEB\sim\Delta AFC(g.g)$
b.Tương tự câu a ta chứng minh được:
$\Delta BAD\sim\Delta BCF(g.g)$
$\to\dfrac{BA}{BC}=\dfrac{BD}{BF}$
$\to\dfrac{BD}{BA}=\dfrac{BF}{BC}$
$\to\Delta BFD\sim\Delta BCA(c.g.c)$
$\to\dfrac{FD}{AC}=\dfrac{BD}{AB}$
$\to AB.DF=AC.BD$
Mà $\widehat{ADC}=\widehat{HDB}=90^o$
$\widehat{BHD}=90^o-\widehat{HBD}=90^o-\widehat{EBC}=\widehat{ECB}=\widehat{ACD}$
$\to\Delta ADC\sim\Delta BDH(g.g)$
$\to\dfrac{AD}{BD}=\dfrac{AC}{BH}$
$\to AD.HB=AC.BD$
$\to AB.DF=AD.HB$
