Giải thích các bước giải:

a.Xét $\Delta BEH,\Delta CDH$ có:

$\hat E=\hat D(=90^o)$ 

$\widehat{EHB}=\widehat{DHC}$

$\to \Delta BEH\sim\Delta CDH(g.g)$

b.Từ a $\to \dfrac{HE}{HD}=\dfrac{HB}{HC}$

Mà $\widehat{EHD}=\widehat{BHC}$

$\to \Delta HDE\sim\Delta HCB(c.g.c)$

$\to \widehat{HDE}=\widehat{HCB}$

c.Gọi $AH\cap BC=F$

Vì $BD\cap CE=H\to H$ là trực tâm $\Delta ABC$

$\to AF\perp BC$

Xét $\Delta BFA,\Delta BEC$ có:
Chung $\hat B$

$\hat F=\hat E(=90^o)$

$\to \Delta BFA\sim\Delta BEC(g.g)$

$\to \dfrac{BF}{EB}=\dfrac{BA}{BC}$

$\to BE\cdot BA=BF\cdot BC$

Tương tự $CD\cdot CA=CF\cdot CB$

$\to BE\cdot BA+CD\cdot CA=BF\cdot BC+CF\cdot BC=BC^2$