Giải thích các bước giải:
1.Xét $\Delta ABE,\Delta ACF$ có:
chung $\hat A$
$\widehat{AEB}=\widehat{AFC}(=90^o)$
$\to\Delta ABE\sim\Delta ACF(g.g)$
$\to \dfrac{AB}{AC}=\dfrac{AE}{AF}$
$\to AF.AB=AE.AC$
2.Từ câu a $\to \dfrac{AF}{AC}=\dfrac{AE}{AB}$
Mà $\widehat{FAE}=\widehat{BAC}$
$\to\Delta AEF\sim\Delta ABC(c.g.c)$
$\to \widehat{AFE}=\widehat{ACB}$
3.Xét $\Delta HBF,\Delta HEC$ có;
$\widehat{FHB}=\widehat{EHC}$
$\widehat{HFB}=\widehat{HEC}(=90^o)$
$\to\Delta HBF\sim\Delta HEC(g.g)$
$\to\dfrac{HB}{HE}=\dfrac{HF}{HC}$
$\to \dfrac{HB}{HF}=\dfrac{HE}{HC}$
Mà $\widehat{FHE}=\widehat{BHC}$
$\to\Delta HEF\sim\Delta HCB(c.g.c)$
4.Xét $\Delta BFC, \Delta BDA$ có:
Chung $\hat B$
$\widehat{BFC}=\widehat{BDA}(=90^o)$
$\to\Delta BDA\sim\Delta BFC(g.g)$
$\to \dfrac{BD}{BF}=\dfrac{BA}{BC}$
$\to BF.BA=BD.BC$
Tương tự $CE.CA=CD.CB$
$\to AB.BF+AC.CE=BD.BC+CD.CB=BC^2$