Giải thích các bước giải:
a.Xét $\Delta ABD,\Delta ACE$ có:
Chung $\hat A$
$\widehat{ADB}=\widehat{AEC}(=90^o)$
$\to\Delta ABD\sim\Delta ACE(g.g)$
$\to\dfrac{AB}{AC}=\dfrac{AD}{AE}$
$\to AB.AE=AD.AC$
b. Xét $\Delta ADE,\Delta ABC$ có:
Chung $\hat A$
$\dfrac{AE}{AC}=\dfrac{AD}{AB}$ vì $AB.AE=AD.AC$
$\to\Delta ADE\sim\Delta ABC(g.g)$
$\to \widehat{AED}=\widehat{ACB}$
c.Kẻ $HF\perp BC$
Xét $\Delta CHF, \Delta CEB$ có:
Chung $\hat C$
$\widehat{CFH}=\widehat{CEB}(=90^o)$
$\to\Delta CFH\sim\Delta CEB(g.g)$
$\to\dfrac{CF}{CE}=\dfrac{CH}{CB}$
$\to CE.CH=CF.CB$
Tương tự $BH.BD=BF.BC$
$\to BH.BD+CH.CE=BF.BC+CF.CB=BC^2$