Giải thích các bước giải:

a.Xét $\Delta ABD,\Delta ACE$ có:

Chung $\hat A$

$\widehat{ADB}=\widehat{AEC}(=90^o)$

$\to \Delta ADB\sim\Delta AEC(g.g)$

$\to \dfrac{AD}{AE}=\dfrac{AB}{AC}\to \dfrac{AD}{AB}=\dfrac{AE}{AC}$

Mà $\widehat{EAD}=\widehat{BAC}$

$\to \Delta ADE\sim\Delta ABC(c.g.c)$

b.Gọi $AH\cap BC=F$

Ta có: $BD\perp AC, CE\perp AB, BD\cap CE=H\to H$ là trực tâm $\Delta ABC$

$\to AH\perp BC=F$

Xét $\Delta BDF,\Delta BCD$ có:

Chung $\hat B$ 

$\widehat{BFH}=\widehat{BDC}(=90^o)$

$\to\Delta BFH\sim\Delta BDC(g.g)$

$\to \dfrac{BF}{BD}=\dfrac{BH}{BC}$

$\to BH\cdot BD=BF\cdot BC$

Tương tự $CH\cdot CE=CF\cdot CB$

$\to BH\cdot BD+CH\cdot CE=BF\cdot BC+CF\cdot BC=(BF+FC)\cdot BC=BC\cdot BC+BC^2$

c.Xét $\Delta DHC,\Delta DBA$ có:

$\widehat{HDC}=\widehat{ADB}(=90^o)$

$\widehat{DCH}=\widehat{ACE}=90^o-\widehat{EAC}=90^o-\widehat{BAD}=\widehat{ABD}$

$\to \Delta DHC\sim\Delta DAB(g.g)$

$\to \dfrac{DH}{DA}=\dfrac{DC}{DB}$

$\to DH\cdot DB=DA\cdot DC\le \dfrac14(DA+DC)^2=\dfrac14AC^2$